The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031239480 Cornell University Library arV19575 Differential equations ,. 3 1924 031 239 480 olin.anx Differential Equations A SHORT COURSE FOR ENGINEERING STUDENTS. By James E. Boyd. Ohio State University, Columbus, Ohio. Published by the Author. 1905. Copyright, 1905, by James E. Boyd. PRICE 60 CENTS. PREFACE. The purpose of this book is to present those portions of Differential Equations which are most used by engineers. It was written with reference to the needs of two classes of students: those who take the subject as a preparation for the study of Mechanics, and students in Electrical Engineer- ing who are already well grounded in the physical ideas in- volved in the electrical problems. The former take the first three chapters, omitting the electrical problems ; the latter complete the book. I am indebted to Edwards' Integral Calculus, and still more to the text-books of W. W. Johnson and D. A. Mur- ray. The symbols and equations of the electrical problems are practically the same as in Bedell and Crehore's Alternat- ing Currents. In my own classes the books will be used to- gether. The examples were prepared with the double purpose of illustrating the text, and of affording a review of some of the more important opertions of the Calculus. Answers are given and stress is laid on the importance of verifying results. In the problems, the physical meaning and use of the arbitrary constants are kept prominent. I am under obligations to Dr. H. W. Kuhn for sugges- tions and corrections. CONTENTS I. DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND DEGREE. ARTICLES. PAGE. 1-3. Definitions 1 4. Integration constants 2 5. Variables separable 3 6. Formation of differential equation.^ 4 7. Homogeneous equations 4 8-9. Non-homogeneous equations of the first degree 5 10-11. Exact differential equations 7 12. Equations which may be rendered exact 9 , 13. Linear equations 11 14. Equations which may be transformed into linears 14 15. Geometrical problems 15 16. Direct current electrical problems 16 17. Alternating current problems 19 II. EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREES. 19. Equations solvable for -^ 23 dx 20. Equations solvable for y 24 21. Equations solvable for x 25 22. Clairaut's equation 26 III. DIFFERENTIAL EQUATIONS OF ORDER HIGHER THAN THE FIRST. 23. Equations with y wanting 29 24. Equations with x wanting 31 25. Applications to mechanics 30 IV. LINEAR DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT COEFFICIENTS. 29. The operator IP 34 30. Solution by means of inverse operators 35 vi CONTENTS. ARTICLES. PAGE. 31. The complementary function 36 32. Case when f(J)) has a pair of equal roots 36 33. Case when f(_D) has imaginary roots 38 34. The particular integral 38 35-36. The inverse operator 39 37. Inverse operators by partial fractions 40 39. The particular integral when ^(^r) is c<^^ 41 40. Case when f{_x') is sina;tr or cosc^i: , 43 41. Case when f(^) is J^w 46 42. Case when /(;ir) is ^wi^F 47 43-44. Problems in mechanics 48 45. Electrical problems 51 ■'i DIFFERENTIAL EQUATIONS. CHAPTER I. DIFFERENTIAL EQUATIONS DEFINED. 1. A differential equation between two or more variables is- any equation containing differentials of the variables. dy = f(xjdx. dy= (x''—a^)dx, Uiy) dy = f {.X , a. X, y'dy-x'dx, U{x, y)dy = f{x)dx, (^ + f)dy = log xdx, are some examples of differential equations. 2. The order of a differential equation is determined by that of the highest derivative in it, and its degree is the highest degree of its derivatives. d''y _i^ dy _y dx'' dx is an equation of the second order and the first degree, while { JyJV\dy_ XTd^) +~SF is one of the first order and second degree. 3. The ordinary integral calculus deals with the differential equation, in which the derivative is expressed explicitly as a func- tion of one of the variables, which may- be regarded as the inde- pendent variable. In the equations dy=^(x' — a^)dx, dy^(x)dx, X may be taken as the independent variable, and we have in the first and, in the second ax In each case the derivative is a function of .*■ only. In like manner the equation dy = fiy)dx may be written dx _ 1 dy /(y)' 2 INTEGRATION CONSTANTS. in which case the derivative with respect to y as the independent variable is a function of y only. In the same way, the equation is a problem of ordinary calculus, and may be solved by integrating twice with x as the independent variable. On the other hand, the equation can not be integrated directly, since y can not be taken as the in- dependent variable. INTEGRATION CONSTANTS. 4. Since the derivative of a constant is zero, a constant may vanish each time an expression is differentiated. Hence, when per- forming the reverse operation, a constant must be introduced on integrating an expression of the first order ; two constants, on integrating an expression of the second order, and so forth. We really introduce a constant in the ordinary integral calculus, when we integrate between definite limits, but it vanishes in the process of subtracting the value of the expression at the lower limit from that at the upper. The value of a constant of integration depends upon the condi- tions of the problem in which it occurs. In a general solution, it should be of such form that it mav have any value from +ooto — oo. Generally, it is expressed as an algebraic quantity as C, or, if the result is more symmetrical as C or C, etc. Soihetitaes it is more convenient to give it some other form, as log C, tan C, tan— 'C, €tc., which fulfill the condition that they may have any valu^ from -|- 00 to — 00. Such an expression as sin C can not be used unless it is known that the conditions of the problem are such that the constant must lie between + 1 and — 1, or unless C is imaginary. In the differential equation xdx + ydy ■=■ 0, the solution is evidently x^ , y -c which may be written more symmetrically Considered geometrically, this is the equation of a circle of radius C. Consider the differential equation of the motion of a body hav- ing a uniform acceleration g. This may be written l^-" -rr dt ^' DIFFERENTIAL EQUATIONS. 3 from which v = gt + C. The constant C, in this case, is the value of the velocity when f = 0, in other words, it is the initial, velocity. A constant of integration, or, as it is generally called, an ar- bitrary constant, has a definite value in a particular problem, but may have an entirely different value in another similar one. The integral of a differential equation with the number of arbi- trary constants corVesponding to the order of the equation is called the general solution, complete integral, or complete primitive. EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE. VARIABLES SEPARABLE. 5. The simplest form of a differential equation is one in which the variables may be separated, so as to bring it into the form f(x)dx^f(y)dy, which may then be integrated directly. For example : (x + a)dy={^+ b)dx. This may be written dy _ dx y-\-b x+a The integral of which is log(^y + b)=log{x + a)+C'; or log(y + 6) =log(jir + o) +log C; whence y = C(.x^a)—b. Examples. 1. -^=J^, Sy' = ix' + C. X^ jS 2. J:L^=x'^y\ A:»y+Cjr + 3 = 0. dx 3. M-=ydx, y^Ce 2 X -^ 4. xydy={:f-^\){.x — \) dx, , x^{y'+l)=:C^^. 5. J^=secydx, sin y=^+C. X 2 6. ^ =Jy-, tan y=^+C. sec'y dx 2 7. (.1+y^) dx=(.l + x^) dy, tan-ij'=tan-i;ir+tan-lC; x-i-C otherwise, y= '^ ^ 4 FORMATION OF A DIFFERENTIAL EQUATION. 8. sjl—y^dx-^l—x^dy, y=X'^l—C^+Cii\.—x\ otherwise, C.i=xi{i.—y^—y\ll—xK y— 1 ar»-3r— 4' ^F+l \x+\/ ' FORMATION OF A DIFFERENTIAL EQUATION. 6. A differential equation may be formed from any equation containing a single arbitrary constant by differentiating once and eliminating the constant between the result and the original equa- tion. If the equation pontains two arbitrary constants, differentiate twice, and so forth. Take the answer of problem 2 of the pre- ceding list of examples x'y + Cy + i^a. (1> Differentiating once we get ix'y dx+(.x^ + C)dy = 0. (2> Eliminating C- between equations (1) and (2) we get ^^x^y\ (3> dx which is a differential equation free from the arbitrary constant C. If the constant stands alone, it vanishes on differentiation, so that it is not necessary to eliminate. For instance, divide equation (1) above by y, and we have y which differentiated gives ?,x^dx—\dy=V), a differential equation free from the arbitrary constant C. Examples. Check the results of the examples of Art. 6 by the method given above. HOMOGENEOUS EQUATIONS. 7. Any equation of the first order and first degree, which is homogenous in x and y may be put into the separable form by sub- stituting vx = y, or vy =: x, where » is a variable. Take the equa- tion (^ + 31=) dx + xydy = (i (1) Let y =: vx, hence dy^v dx-\- x dv. Substituting in (1) x'd+v') dx + x'vivdx + xdv) (2) Dividing by .*• and collecting (l + 2z/')dx + vxdv — 0. ' (3) DIFFERENTIAL EQUATIONS. 5 This may be separated into dx j_ vdv _^ ,4, Log ;^+ A.log(l +22-2) =log C. ( 5 ) jr*(l + 2l'')=C' (6) Substituting z/=_2Lwe get r' + 2;try == C (7) The general homogenous equation is of the form (Axn -\-Bxn-^y-\-Cx^-^y^ .. ..^Ryn)dx -^(A'xn +S'x''-'y + C'^n-V +J?'y'')ay=0. Substituting 31 ^ vx, dy = vdx -\-xdv, we get x''(A+Bv+Cv^ .. -\-Rv")dx +X1 (A'-\-B'v-{-C'v^ +R'v«)(vdx+xdv)-fi. Dividing by x^ and collecting iA+{A'+B)v+{B'+C)v^ +R'v''+i]dx +{A'+B'v+C'v'' +R'v")xdv=0. dx , A--\-B'v^-C'vi ^R'yn ^^^^ X A+{A'+B.v+{B'+,C)v^ +R'v"+i From this it is evident that any homogenous equation is rendered separable by substituting y = vx, and, from the symmetry of the terms, the same is true for x =^ vy. Examples. 1. {y'—2xy+2x')dy=iy'dx, y—x=C{f—2xy). 2. {x+i^y^+xy) dy=ydx, Cy=e^ ^Z-^ - " y 3. iy+i^x^+y^) dx=xdy, Cx^=y+isjx^+y', Otherwise, CV — 2Cy = 1. 4. Check the answers of the above problems by differentiating and eliminating the constant of integration. NON-HOMOGENEOUS EQUATIONS OF THE FIRST DEGREE IN X AND Y. 8. Equations of the first degree in x and y which are not homogenous 'are of the form (ax-\- by -\- c)dx -\- (a'x-\- b'y-\- c')dy =^0. (1) Such equations may generally be transformed to the form (ax + by')dx'-\- (.a'x'-\- b y')dy '= C, (2) which is homogenous and may be solved bj the methods of the preceding article. In equation '1) let x = If' -\- h, J' = y' -\- k, hence dx = dx', and dj* — d/ , (3) where h and k are undetermined constants. 6 NON-HOMOGENEOUS EQUATIONS. Equation (1) then becomes {ax'-\- by'+ ah-\-bk + c)dx'-{- (ov+6y+o';j + &'fe + c')rfy = o. (4) If now we assign to h and k such values that \ (5) a'h + b'k + c'= J Equation (4) becomes (2). It is necessary, then, to solve the homogenous equation (2), which, neglecting the prime, is the same as (1) with the con- stants omitted, and substitute from (3) x' =^ x — h, y' =: ji — k. Take the equation {x + 2y + \)dx-{x — l)dy=0, (1) Substituting and dropping the primes {x+iy)dx — xdy = (i, (2) which solved as a homogenous equation gives Cs^^y + x. (3) For the constants h and k we have ;i + 2fe + i = o, (4) h — \ = 0, Whence fe := 1, and k = — 1. Substituting x — 1 for x and ji + 1 for ji in equation (3) we get C{x-\y = y^-x. Examples. 1. {2x + y + l)dx+{x + 2y — 4:)dy=(i, x' + xy + y' + x — iy — C. 2. {x + 2y + S)dx=(,2x + y + S)dy, C(x-yr = x + y+2. 3. (.4x — y + 5)dx+(x — Ay + 5)dy = 0, ix-y + 2r(^ + yy=C. 4. (x-y)dx+(x + l)dy^O, x+l^Ce-^ 9. If in a differential equation of this kind, — ;— = .. the a ty method given above fails, for the equations for finding h and k are of the form a/i + 6fe + c = 0, rah + rbk + c' = 0, the loci of which represent parallel straight lines when c' is not equal to re, and the same straight line when c' = re. In the first case, the values of h and k are infinite, and in the second case, they DIFFERENTIAL EQUATIONS. ' T are indeterminate. These are easily solved, however, by a dif- ferent method. For the indeterminate case divide by {ax-\-by-\-c) . Take the equation {2x — ^y + 6)dx—{ix — Qy + Q)dy = 0. Dividing and transposing, we have 2dx — My, Hence 2^ = Sy + C. For the first case, let ax -^ by =^ z, a new variable. The method is shown by the problem below : {x + y + 2)dx-\-{2x + 2y + l)dy = (i. (1) Let x-\-y^z, hence dx^dz — dy. ' {z + 2){dz — dy) + {2z + l)dy = 0. (2) {z + 2)dz+{z — \)dy = 0, (3) ^ + 3' + 3 1og(a-l)=C, (4) ■ ^ + 23. + 3 log(x + 3--l)=C. (5) Examples. 1. (.x-^y)dx—{x + y+l) dy = 0, 2x — 2y + C^log(2x + 2y + l). 2. {2x + y + S)dx+iQx + Zy+Q)dy-0, x + iy = C. 3. {x + 2y-\-Z)dx— (2x + Ay+l)dy = (i, x — 2y + loz{ix + %y + l)^ =C. Miscellaneous Examples. 1. {x+\)dy=z{y + 2)dx, C{x-\-\) =y-\-2. x — l 2. {x-\-y')dy={y + l) dx, C+Cy^eT+l dx VI— jr2 ' 4. {x + y)dx+{2x + 2y + \)dy = (i, \osC{x-]-y+\) = 2y + x. ' EXACT DIFFERENTIAL EQUATIONS. 10. An exact differential equation is one which is formed by dif- ierentiation alone without elimination of constants, or reduction. X dy -\- y dx ^ a is an exact differential equation, being formed by differentiating xy = C. If we differentiate f{x.y)=C. we get 8 EXACT DIFFERENTIAL EQUATIONS. -^f(x,y)dx^^f(x, y)cfy=0 dx dy as the standard form of an exact differential equation. This may be written Mdx + Ndy = 0, where M is the partial derivative of the function with respect to jc and N is the partial derivative with respect to y. An equation of the form Mdx + Ndy = a, is exact if dM ^ dN dy dx ' For M=-4-/(x,y), N=-^f(x,y), dx dy that is: — = — is obtained by differentiating f(x,y) with respect dy ts X and then with respect to y; while -^ — is obtained from the same function by differentiating with respect to y and then with respect to x, and as it makes no difference in what order these operations are performed, the expressions are equal. To il- lustrate, take the equation x'y + 3xy = C, differentiating ^2xy + 3y)dx + (x' + Bx)dy = 0, an exact differential equation. Differentiating (2;rj» + 33i) with respect to y we get -^(2xy+3y)^2x+5=^. dy dy Differentiating (.v" + 3x} with respect to x, ^(;t2+3.r)=2.r+3=4^. dx dx SOLUTION OF AN EXACT DIFFERENTIAL EQUATION. 11. Integrate M dx, regarding y as, a constant and ;ir as a var- iable, then integrate the terms of N dy which do not contain x, regarding ji as a variable, and add the results. For example we will solve the differential equation (■i:^ + '^xy)dx+ {'2x'-\-2y)dy = 0. This is an exact differential equation for -^{3x^+ixv)'^ix, and -^{2x^+2y)=ix. dy ' dx DIFFERENTIAL EQUATIONS. 9 Integrating {^%x^-\-^xy)dx, regarding x as variable, we get The only term of i^K? -\-'iy)dy which does not contain x is I'^dy, the ihtegral of which is y^ Adding, we have for our solution ^ + 'ix-'y + f = C, which may be verified by differentiation. The reason for this method of integration is evident from the following considerations : Any function of *■ and y is made up of three seits of terms : (a) those containing x only, (b) those containing y only, and (c) those containing both x and y. On differentiating to form an exact differential equation (a) pro- duces terms of Mdx which, of course, do not contain y, (b) pro- duces terms of Ndy which do not contain x, while (c) produces terms of Mdx which contain y and terms of Ndy which con- tain X.. If we integrate the terms of Mdx which do not contain y we get the terms which contain x only. If we integrate the terms of Ndy which do not contain x we get the terms which con- tain y only, while if we integrate either the terms of Mdx which do contain y or the terms of Ndy which contain x we get the terms in X and y. Examples. Show that each of the following differential equations is exact and solve. 1. i'lxy'' -\-1y)dx+ (3.*-y -|- 2jr -|- 231) dy = 0, xy + 2xy + / = C. 2. 2xdx + 6xy dy + By'dx — 0, r'-\-Sxy' — C. o dx _ xdy jdx _(. ^,x_^ ■ y j,2 "^ 2 ' T" "2~~ xdx-\-ydy dx ^" v^^Hy '^^'^°' v^^-^-y^-log ^=c 5. yx^-^dx-irxy \oz xdy=^, x^ = C. EQUATIONS WHICH MAY BE RENDERED EXACT. 12. Sometimes an equation may be made exact by multiply- ing by a factor. The equation sf'dx -{-xy dy^O is not exact, but if it is multiplied by— —it becomes X dx -\- y dy =^ 0, which is exact. The factor which renders an equation exact is 10 EQUATIONS WHICH MAY BE RENDERED EXACT. called an integrating factor. Integrating factors may frequently be found by inspection. There are a number of methods for finding integrating factors. We will give one which is applicable to cases where the factor is a function of x only or of ji only. [For other methods, see Mur- ray's Differential Equations.] Let the equation be W dx + W dy=.^, and let R be the integrating factor. The equation RW dx-\-RW dy — 0, is exact, and ^(RM')=-^{RN). (1^ dy dx Suppose i? is a function of x only. Since i^^O, and i^=^^ dy dx dx j^dM-_ RdN' _^ NdR ^2> dy dx dx dR _ \ / dM' _dN' \, 3 If R is, as we have assumed, a function of x only, the coefficient of dx in (3) is a function of x and is directly integrable. Take the differential equation (atV — y)dx+ (x'y + x)dy = 0, Assuming that i? is a function of x we have from (3) dR _ {2x»y—l—4x«y—l)dx __ 2dx R x*y-\-x X Hence, R= is the integrating factor. x^ If the expression 1 /dM'_dN'\ N\ dy dx )' is not a function of x, ive know that the assumption that i? is a function of x only, is incorrect. We may then assume that R is a function of y, and we get from (1) As the expression on the parenthesis in equations (3) and (4) is the same, the work of finding the integrating factor, wheij it is a function of one variable only, is not great. Take as another ex- ample. ^3x^y^—JL\ dx+{x»y+l)dy=0, dy dx X DIFFERENTIAL EQUATIONS. 11 If we attempt to divide the term to the right of the equality sign by the coefficient of iy, we find that the result is not a function of X only, consequently the, integrating factor is not a function of X. The term is divisible by the coefficient' of dx, the quotient being 1 y ' Hence from (4) 1^=-Jy-, and J- R y y is the integrating factor. The exact differential equation -is (%x'^y——\dx-\-(x^-\^ 2—\ dy =0. Examples. Find integrating factors, solve the equations and verify the results. tan^= <^-y . >2 1 + Cji 2. (3xy — ?>y^%x-\-X)dx^{x—\ydy = ^, r + ^+1 C x—\ (^—1)8 3. %x^dx + xdy — ?,ydx = <). y = Cx' — 3x*. LINEAR EQUATIONS. 13. A linear differential equation is one in which the dependent variable and its derivatives are of the first degree. The standard form of linear differential equation of the first order where y is the dependent variable and x the independent variable, is ax P and Q being functions of x only. This equation may always be made exact. To prove this statement and find the integrating factor, arrange the terms and multiply by R, which is the assumed integrating factor. R{Py — Q)dx + Rdy — 0. (1) This equation being exact ^R(Py-Q)=^ (2) dy dx If /? is a function of x only (2) becomes RP=.^, dx 12 LINEAR EQUATIONS. R iPdx Hence i? is a function of x only and is equal to « ' Substituting this value of R in (1) and rearranging the terms in the order of the standard form, we have eV^'^dy + Pye% ''^''dx = Qe^ ''^^dx. (3) The derivative with respect to .v of the coefficient of dy is Pe ' , and the derivative with respect to y of tile coefficient of dx is also Pe ' , consequently equation (3) is exact. To solve (3) integrate the coefficient of dy, regarding y as the variable, and those terms of the coefficient of dx, which do not contain y, regarding x as the variable. The final result is yeV'=^Qe V^dx + C. (4) The differential of the first term of (4) gives the first two terms of (3), while the differential of the two terms to the right of .thp equality sign of (4) is the last term of (3). Take as an example, J^+JL=x\ (1) dx X jPdx =J ^^ -log X, e > =e Equation (3) becomes xdy -\- ydx ^: x^dx, the solution of which ,is X* „ x^ r xy=^+C, or y=f^+l-. 4 ^ X When y is an explicit function of x and of the first degree, it is frequently convenient to verify the result by substituting this value of y in the original equation. For instance, if we substitue in equation (1) above the value 1- — =y, we get 4 ■ X ^+JL=^-^+^+^^x'. dx X i x' 4t x' Examples. 1. ^+2xy=4x, y=2—Ce-^. dx - 2. ^+J?L=5 (.;!-+ 1)2, y^(x^\)^^. ^ -- dx x+1 i.x+l)^ -^ — y=a, y=Cex — a. dx DIFFERENTIAL EQUATIONS. 13 4. ^-\-y=x, y=x-\-\-Ce-x. dx 5. ^+3y=^, y=\^^\.-^Ce-^ ax 3 9 6. -^+v=sin jr, r= (sin ^ — cosx)-{-Ce—x. dx 2 7. :*i+a^=:sin «^, ^==^ii^i£=*?£!j£+C^-, To integrate e"^ sin 6;ir d;ir^ we make use of the method of integration by parts. Since d(uv) ^udv-\-vdu I udv — uv — I vdu. If we let e<^x dx = dv and sin hx = u, this becomes Jgax sin i;jr(/;ir = — -ea^ sin i;ir — ( e"^ cos bxdx. ( 1 ) a a J We may operate in the same way on j e''-^ cos bxdx and get as the final term — I e^^ sin bxdx. As this is the same integral as the one with which we began, it may be transposed to the left of the equality sign and added. The solution will then be obtained by , , dividing by *^ „ — . Another method is to integrate the original ex- pression by parts, letting sin bxdx=dv and e<'x=u. This gives j e<^x sin bxdx= — —r-e"^ cos hx +-^ fea^ cos bxdx. ( 2 ) Multiplying (1) by a and (2) by h and adding (a^-\-b^ j if^siu bxdx=e<"^ (a sin bx—b cos bx). To put the answer into the second form requires a trigonometric transformation; a sin bx — b cos bx may be put into the form of the sine of the difference of two angles by dividing by a factor r such that ^ equals the cosine, and — . equals the sine of the same angle. r r To find the value of r, « sin 5;i; — bcosbx_/ a , b \ 1 ,— ; — Ts 1 — =^=: sin bx — • ^=: cos bx 1 =, from which the second form of the answer is obtained. 14 EQUATIONS TRANSFORMED INTO LINEAR. Examples. 8. Sin>'-^+;r-lcos>=— jr! Co% y=^-\-Cx. dx 2 9 -^+losy = x^, \og y-x^—2x+2—Ce-x . ydx 10. -^+^-^=y, ^(j+l)=Zl+^+C dy y -^-1 ■' ' 4 3 11- dy^-===dx = xdx, yi,xJr<^^T^'')=-Y-^^^^-^^^-^C. EQUATIONS WHICH MAY BE TRANSFORMED INTO LINEARS. 14. An equation of the form ^^Py = Qyn, dx is a linear. If we divide by y" it becomes y-"-^+Py^-^^Q, dx which is equivalent to 1 d -(y^-'>)+Py^-»=Q. 1 — n dx Multiplying by (1 — n) this is dx a linear in which y^ — " replaces the y of the standard form. As an example dx^ -^ -^ y-^^+xy-^^ix, ax dx _ x^ £2_ (i+Ce 2 )j,/— 1=0. Examples. 1. -^+J/ tan ;t-=i55^, ' y^=l+Ccos^x. dx y ' _^_2tanj/^ , siuj,_ (^+l)=^ dx x^-l ■^' ;t- — 1 3 ^ (7 jr+1 DIFFERENTIAL EQUATIONS. 15 rf;tr J- ' 5 Miscellaneous Examples. Solve and verify results : 2. xdy-\- 2y dx = ^'£i;r, 6y = ^'' + C'x— ^ . ■3. (.r + 3)) d^+ (2;ir + 23/ — 3)£;3/ = 0, ;t: + 2j + 3 log(*- + ji — 3) = C. ■4. 2yd;r — (A-= + 2;f3; — y')dy^<), y{x-\-y) ^C{x — y). ■6. y''dx + ;f y rfji + 4 rf;ir = 0, j/ = JL + 2 tan - 1-^+ C. 6. 3 cfj) + ji tan ;i: d;r = ji— ^cotan X dx, y^sec X = log tan-^+ C. 7. xdx + V -^^ + y^ rf;ir+>'rfj' = 0, :v + V^^ +y = C". 8. _^+f^j' = g^, y = l+Ce-e'^ dx 9. (43)+^')d;i: = ;«:£;3;, y = C;f- — ;i:l 10. (l&n y + — - — ^dx + -^ s^c^ydy = 0, x tan j/+tan- 1 x=:C. 15. 1. Find the equation of the curve for which the tangent at any point is perpendicular to the radius vector. The slope of the tangent to any curve is_2L. The tangent of dx the angle which the radius vector makes with the X axis is-i- X Hence the differential equation is dy X dx y ' the solution of which is a circle of radius C. The constant C may have any value from to 00, unless limited by conditions other than those stated above. The result means that in any circle, whatever the radius, the tangent is perpendicular to the radius. 2. Find the curve for which the tangent of the angle which the radius vector makes with the x axis is twice the slope of the tangent to the curve. The parabolas y^ = Cx. 16 DIRECT CURRENT ELECTRICAL PROBLEMS. 3. Find the curve for which the tangent and the radius vector make supplementary angles with the x axis. Any rectangular hyperbola of the form xy = C. SOME DIRECT CURRENT ELECTRICAL PROBLEMS. 16. In these problems we will use the following symbols : C capacity, R resistance, L inductance, q quantity at any instant, t current at any instant, e potential difference at any instant, e base of natural logarithm, Q maximum quantity, / maximum current, E maximum E.M.F. or potential difference. 1. The terminals of a charged condenser are connected through a non-inductive resistance. Find the expression for the charge at any instant. e , . 1 The current strength is— and e = -^. The current is also lal to the rate of d differential equation equal to the rate of discharge, that is, i = — —2-, hence, we have the at -1-=-^ (1) RC dt * ' Equation (1) is the case of variables separable, and its solution is i_ g=C,e ''^. (2) The constant Ci is obtained from the condition that when t = 0, t_ g= i5, and consequently c =1. Hence, Cj = Q, and (2) be- comes t_ q-=Qe . (3) Equation (3) may be written A condenser of 2 micro-farads capacity is discharged through a resistance of 100,000 ohms. What portion of the charge remains in the condenser after .1 sec, .2 sec, .5 sec, 1 sec, and 2 seconds? DIFFERENTIAL EQUATIONS. 17 In this problem, /?C = 10^ X 2 X 10-° - .2, and -^ - 5. The work may be conveniently arranged as shown below : Time. IvOg«- Q_ q I-ogiof ^°^vr^ g Q .1 .5 .21715 —1.78285 .6065. .2 1.0 .43429 —1.56571 .3679 .5 2.5 1.08573 —2.91427 .0821 1.0 5.0 2.17147 —3.82858 .0067 2.0 10.0 4.34294 —5.65706 .0000 Show that the charge in the condenser decreases in a geometricar progression. If we can measure a quantity equal to one ten-thousandth of the original charge, how long will it take to discharge this con- denser completely as far as our measurements will enable us to determine? 1.842 seconds. 2. A condenser is connected suddenly through a non-inductive resistance to a source of constant E.M.F.. Find the charge at any time t. If E is the E.M.F. of the source, the effective potential dif- ference at any instant is £ — e, hence E-e = Ri=R% (1) at ^naE—^=R±, (2) C ai is the differential equation required. The solution of (2) is / q—CE=C^e ^ . (3> Substituting for CE its value Q, and inserting the condition that g = when * = 0, to get the arbitrary constant Cj, we find as the final result =X-r"'0. (4> 3. In problems (1) and (2) find e and i. t From (\)e = Ee , t E t t i = —^Q e RC —— j^ e Rc = —le rc From (2)e=.£:( \-e "^J, 2 D B 18 DIRECT CURRENT ELECTRICAL PROBLEMS. E L- i = j^ e RC . What is the meaning of the negative sign before the first ex- pression for current? What assumption in the derivation of the formula determines this sign? 4. The terminals of a constant potential dynamo are suddenly connected through an inductive resistance. Find the expression for the current in terms of the time, the resistance of the entire circuit, and the inductance of the entire circuit. The E.M.F. must be equal to the drop of potential due to the resistance, together with the counter E.M.F. due to inductance. Hence we have the physical equation ^ = ^^' + ^. (1) Hence, dividing by L we get the standard linear form of which the solution is E Rt i^-^ + C-^e L , (3) The constant Ci is determined from the fact that when ; ^ 0, i =0, hence, C^ — — -— = — /. The final equation is then R i = l(l-e~^). 5. The terminals of a coil of inductance L, in which a current / is flowing, are suddenly joined together through a non-inductive resistance and, at the same instant, the battery supplying the cur- rent is disconnected. Determine the current after an interval of t second. Ri t = le , R being the sum of the resistances of the circuit. If the inductance is 20 miHi-henrys, the resistance of the coil 2 ■ohms, the non-inductive resistance 98 ohms, and the initial potential ■difference 2 volts, find the current and potential difference after .0001 sec, .0005 sec, and .001 sec i .606 .0821 and .007 amp. e 59.44 8.04 and .66 volts. 6. A cell having an E.M.F. of 2 volts and negligible internal resistance is connected to a cell of 1 ohm resistance and .01 henry DIFFERENTIAL EQUATIONS. 19 inductance. After the current has became steady 9 ohms non-induc- tive resistance is suddenly cut into the circuit. Derive an expres- sion for the current, and compute the current, the potential dif- ference of the terminals of the cell and of the terminals of the non-inductive resistance after .0001 sec. 7 = .2 -h l.Se-™"'- Potential difference of terminals of coil 16.29 volts. ALTERNATING CURItENT PROBLEMS. 17. The typical form of an alternating E. M. F. is expressed ■e^E sin i^t, when w is equal to 2t times the number of complete periods per second, and t is measured from a position in which the ■e is passing through zero in the positive direction. 1. Circuits having resistance and inductance. Our differential equation after dividing by L becomes di I R ; E . ,, , 1 . The solution of this linear equation is ie i =^J ^ -^ sin «* d* -I- C. ( 2 ) This equation is the same as Ex. 7 of Art. 13. Performing the integration in the same way we get fit i = ^ ^^ ^'" "^ — ^" '^°^ "^^ A- Ce ^ r S > which reduces to A<^t — tan-^-^]+Ce l, (4) Tan-i— is the angle of lag, and v' i?" + LV is the impedance. R C is the arbitrary constant. We will find the constant C for a numerical problem. Given L = .045 henry, /? = 24 ohms, E — 150 volts and oi 400. From this we get the impedance to be 30. We can not make the assumption that J = when ; = as in the preceding direct current problems, for we have already made the assumption that i^O when e = 0. The time of one complete period is .0137 sec. Curve I of Fig. 1 represents E. M. F. for a little more than two and one-half peri- ods, the abscissas being time and the ordinates electromotive force. One division, representing .002 second, corresponds to 45°, 50' of arc. 20 ALTERNATING CURRENT PROBLEMS. Curve II of the same Fig. represents (sino)* — 0). ViS^' + LV The angle of lag is 36°-52', corresponding to .00161 second of time. This curve first crosses the zero line .00161 to the right of the corresponding point of the E. M. F. curve. ' The constant C depends upon the point where the circuit is closed. In the figure we have pjotted the case when it is closed .003 second after initial time. We liave (w* 0) =3r-53'~= 1.6, and find C«-i« = —2.641, and C = — 2.641«i-6 = —12.93. Jit Having C, we may now plot the curve Ce ^ of Fig. 1. This is curve III r / \ r ^ r 1 1 s / \ \ \ f \ j \ j \ I ^ \ / ^ $ ^ / ^ 1 --s \ JL. f / ^ K '1 l\ \ 1 / \ / / \ .■■■' ^ \ / / \ / 1 \ /-^ \ \ y \ 1/ £ « ^ 1 \ v^ 1 ^ j s / \ / ' ^ \ J § / \ / ra \ \ / 1 K J BCO WW ^ a n a i*OI w o> oaoi B « S .0 4 i? IB iom t Oi #«u BWl aoi u a le th M.a vm H04 Figure 1. In reality we need only that part of the curve beyond t = .003, Rt so we might as well have used Ce L and plotted from A to the right. The current curve IV is the sum of curves II and III. At the instant of closing the circuit, as we have assumed in finding C, the current is zero and curve III is just as far below the zero line as curve II is above. The current curve IV then starts at zero at .003 sec. and lies below curve II until it reaches about .01 sec, where curve III becomes negligible. From that point it coincides with the sine curve II. The effect of the exponential term in this case is appreciable for about the time of one alternation. Rt For given values of R and L the curve Ce L has the same form whatever the value of C. For different times of closing the circuit DIFFERENTIAL EQUATIONS. 21 it is only necessary to displace curve III to the right or left so as to make it pass through the point whose abscissa is the time of closing and whose ordinate is that of curve II with the opposite sign. For instance, suppose that the circuit is closed when t is .006 sec. The ordinate of II is 4.9, and curve III must pass through the point whose co-ordinates are .006 and — 4.9. EXAMPLES. 1. Find the value of C for the following cases : When the cir- cuit is closed at .006 second, .009 second, .012 second, and .015 second. First case, C ^ — 120 Last case, C = 11920. 2. Construct the curve be M5 on a sheet of thin cross-sec- tion paper and construct curve II on a second sheet. Combine the two to find the current curves for the cases when the circuit is closed at .002 sec, .004 sec, .0096 sec, and .012 second. Rt The curve Ce l- represents the fall of a direci ctirrent of ini- tial value C in a circuit of resistance B. and inductance L. We may imagine that, at the time of closing the circuit, the alternating cur- E T&aX—i= sin {yt — 0) begins at once at its full value, but that Vi?^ -|- Ui-r there exists with it a direct current in the opposite direction, which at that instant, is exactly equal to the alternating current, and which dies away as if the alternating current did not exist. Circuits having resistance and capacity. An alternating E. M. F. is suddenly ajiplied to a circuit consist- ing of a condenser in series with a non-inductive resistance. Find the current at any time. The E. M. F. equation is I?i + -^^ E sin i^t, (1) Substituting for i its value--^ we get the linear dt The student will show that the solution of this linear reduces to ? = — 1 _ ^ cosf at + tan -l_-j + C^e'^C ^ (gj wJ7?2- 22 CIRCUITS HA VING RESISTANCE AND CAPACITY. From (3) we get by differentiation ]+Cjff CR, (4) W^^ + ci^H"'^*^""^)" where C,=.-^. Tan — 1 is the angle of advance, and may be represented by Cxtw 0. If the time of closing the circuit be represented by *, we have, since g = 0, ^ ^ <^.* ^^"~ ■ 1 cos (.Of + 0) ^W^ + ci- If u = 400, i? = 500 ohms, C = 10 microfarads, and e = 150 sin at, construct q and i curves the circuit being closed when t .003 sec. CHAPTER II EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREES. It is customary, when dealing with equations of higher degree, to express the derivative, -f-, by the single letter p. An equation ax of higher degree then, is a function of x, y, and p. There are three principal classes of these equations: 1. Equations solvable for p. 2. Equations solvable for y. 3. Equations solvable for x. EQUATIONS SOLVABLE FOR P. 19. When an equation can be solved for p without giving terms in x and y too complicated for integration, this method should, in general, be used, as it gives us one or more equatior(s of the first degree to which the methods of the preceding chapter apply, for example : Solving for p : The solutions of these equations are ^ = 2;^,and^: ax dx y = x^ + C,anAy = ^ + C'. (2) Each of these is a solution of the differential equation. Both may be combined to form a single solution: (j'-C-.r2) (j/-C-^)=0. (3) As this is the solution of a differential equation of the first order we need only one arbitrary constant in the result, and accordingly we drop the prime from the C in the second term. We will show now that (3) is the solution of (1) by perform- ing the operation of differentiating and eliminating the constant. Differentiating (3) and dividing by dx: (y—C—x^) ip—x) + (y—C~^){/> — 2x) = 0. (4) 2(2p — 3x)' ' ' 23 24 EQUATIONS SOLVABLE FOR P. Substituting the value of - (31 — C) in equation (3) and combining we get 4(2/> — 3^)=! whence we get (/> — 2;r) (/> — x) = which is equation (1). EXAMPLES. Solve the following differential equations and verify the an- swers : 1. />= = 4<', (y-Cy=:x\ EQUATIONS SOLVABLE FOR Y. 20. An equation which can not be conveniently solved for p, may often be solved for y, the factors being in the form of y = f(.x,p). (1) To integrate such an expression first differentiate it, and divide by dx. Since — ^ = J>, we get a differential equation which does not contain y. Regarded as an equation in x and p ,{2) is of the first degree. Accordingly it may be solved by the methods of the preceding chapter, giving a result as a function of x, p, and C. The final result is obtained by elim- inating p between this result and the original equation (1). Take the equation 2y = px + p\ (1) Differentiating and dividing by dx 2p^p+{x + S/>^)^, (2) ^-^ = 3A (3) dp p a linear equation, the solution of which is x = 3p'+Cp. (4) we must now eliminate p between (4) and (1). If we attempt to do this by substitution we shall get results involving surds. It is better therefore to use another method. If we multiply (1) by 3 and (4) by p and substract, we get Cp' — ipx + 6y = 0. (5) We have eliminated p^ and have now two equations (4) and (5) of DIFFERENTIAL EQUATIONS. 25 the second degree. We will eliminate p'' between these last two equations and get p(_C''-^nx)^Cx-\-\%y. (6) Substitutifig the value of p from (6) in equation (4) or (5) we get 2Cji= — Ox- + i%Cxy — Vox' + IO831 = 0. EXAMPLES. Solve and verify results. 1. y^-m = 0, \y + Cx- X — 2C;ir>«+C2=0. 2. y + /2 = 0, y (^ - + C )2 + 4 = 0. a y- = p\o%p. . y = e -1 ± 1^ 2;tr + C + 1 (- + 2 -1 + iog/ = 2^ + ay From (3) we get the linear dq _,Zq _ _ J_ / 3 V dy ^ 2y 2y^' gy =y -\-C. ^'' Substituting (4) in (1) we get the same result as in the preceding case. EXAMPLES. 1. ;f^^y+y=0. 4ar + (logO/)« = a 2. x + {^y=0. ix+(y—C)^ = Q. clairaut's equation. 22. An equation of the form y = px+ f{p), (1> is called Clairaut's form. Treating it as an equation solvable for y we get /> = /> + u + r (/'))-''" Equation (3) is satisfied by .«■ + f {p) From the latter we get p = C, which s.ubstituted in (1) gives as the general solution y = ex + fiC). (4) All that is necessary to be done in solving an equation of Clairaut's form is to substitute C for p in the original differential equation. There still remains the other factor of (3), X + fip) = 0, (5> which, regarded as a function of p, and x does not contain any dif- ferentials. A solution may be obtained therefore without any inte- gration by merely eliminating p between (5) and (1). The result thus obtained, since there is no integration, does not contain any arbitrary constants, and is called a singular solution. We have dx' (2> 0. (3> — 0,and dp dx = 0. DIFFERENTIAL EQUATIONS. 27 then two solutions of Clairaut's equation; a singular solution with- out arbitrary constants and a general solution with an arbitrary tonstant. Take as an example .y = xp+^. (1) Differentiating and dividing by dx hence From — S- = 0, we obtain the general solution, ax y=Cx + ^. (4) 4 If we eliminate p between x + S— = and equation (1), we get the singular solution y + J, = 0. (5) If instead of eliminating p we integrate the equation ^ 2dx we get x' + y = C^, (6) which is the same as the singular solution with the addition of the constant Ci. The meaning of these results in a geometrical problem is seen from Fig. 2. The straight lines represent the equation y.= Cx+J^, for values of C differing by .2. For instance, the line 4 — 4 corre- sponds to the case where C is .4, the line 6 — 6, to the case where C is .6, etc. The singular solution, x" -\- y ^ 0, is the equation of the para- bola (not drawn in the figure) to which all lines j; = Cr + ——ate 4 tangent; that is, the singular solution is the envelope of the lines representing the general solution. The equation x' -\- y ^ Ci, represent parabolas with vertices displaced upwards a distance Ci. We will show that any one of these parabolas is the locus of the intersection of lines, y=Cx + ^, 4 28 CLAIRAUT'S EQUATION. for which C differs by a definite amount. In the figure we have ■drawn the parabola for the intersection of the lines for which the values of C differ by unity. One point on this curve is the intersection of the line — for which C is with the line 10^10 for which C is 1. Another point is the intersection of 2 — 2 with 12 — 12. To find the equation of this curve let y =: Cx -\- C^ , 4 y = {C -\- l)x ^ (C + 1)^ be the equations of the intersecting lines. Eliminating C we get x' ^ y =. — -, Show that the locus of the intersection of the straight lines y = Cx + — - for which the values of C differ by a constant amount a is the parabola i:^ + ji = Ci, where Ci = 16 Figure 2. MISCELLANEOUS EXAMPLES. 1. y = px-\- p\ 2. .= ^^-lo.(-^) 3. X ■ dx dy dy ~dx 4. m ■+2. dy 5. Tan-i^L dx dx = .r— 1 General solution, y = Cx -\- C; Singular solution, ^li' + llf = 0. y = Cx — log C; and y = \ -\- log x. ^-Cy+^, or x'" — 4jc = Cj. y + x=C±^. y + log cos ( ^ — 1 ) =! C -r2 + 1 = 0, CHAPTER III. DIFFERENTIAL EQUATIONS OF ORDER HIGHER THAN THE FIRST. 23. We will lake up the following forms of diflferential equa- tions of higher orders: 1. Equations with y wanting, 2. Equations with r wanting, 3. Linear equations with constant coefficients. EQUATIONS WITH Y WANTING. 24. These equations are of the form and are solved by simple integration, a constant being inserted each time an integration is performed. Take the equation Multiplying by dx this becomes •m Hence which integrated again gives 25. This simple form of differential equation has many appli- cations. One of these is the case of a moving body. If j is the ds distance of a body from some point taken as the origin, — =- is the d( d's velocity and 5- is the acceleration. df Problem. A body moves with a uniform acceleration a, find the velocity and position after t seconds. The' constant Ci is the initial velocity, and C2 represents the initial displacement, that is, the position of the body with reference to the origin at zero time. 29 30 EQUATIONS WITH Y WANTING. Another important application is that of a beam with a trans- verse load. The differential equation of an elastic beam is EI^ = M. E being the modulus of elasticity of the material, /, the moment of inertia of the cross-section with respect to the neutral axis, and M the moment of the applied forces with respect to the neutral axis. A beam is fixed horizontally in a wall (Fig. 3), and projects a distance, I, to the right. A load, P, rests on the free end. Find the equation of the elastic curve and the deflection of the free end. A section distant x from the wall is / — x from the free end, and the moment at the section due to the load P is — P (J, — x). The negative sign is used because the slope of the beam-ji becomes ax less, algebraically, as x increases. Hence the differential equation is (1) EI dx^ = -/>(/- -X), r ~3^ V.',;; I I "Z^^"""^^ 'IGURE 3. i dx \ 2 / " (2) To determine the arbitrary constant, Ci, we have the condition that the beam is horizontal at the wall. Hence -4- = when ;r = 0, dx which substituted in equation (2) gives Ci = 0. Integrating again, we get Ely = - P(J^ -^^ + C„ (3) and from the condition that at the wall where ;r = 0, j; = 0, we get Ci =: 0. The equation of the elastic curve is then At the end where x^^l the deflection is ZEI Solve this problem integrating (/ — x)dx as a single quantity. That is, write equation (2) DIFFERENTIAL EQUATIONS. 31 ElJ^ = f-d — xf + C dx 2 Show that the results are the same, but the constants different. e Figure 4. Solve the same problem with the beam projecting to the left of the wall (Fig. 4), taking the origin at O, the position of the free end of the beam before it was deflected. EQUATIONS WITH X WANTING. 26. Take first the equation of the second order (i'y _ dx" f(y) calling — -^ = /> as in Chapter II, this is ax dp dx f(y), (1) To solve multiply dy dy, whence fdp = f{y)dy. This integrates to -^ = ff(y)dy + C. . Equation (4) is of the first order and second degree, and is integrated the second time by the methods of the preceding chapter. (2) (3) (4) Examples. 2. dx' d'y by, < b -c,. dx' - y, X = log {y + yj,2 -I- C2) + Q- or jr = s\uk — 'i-~- -\- C^. '- 1 27. Example (1) has some interesting applications. One of these is the case of mass vibrating on a spring. When the distortion of a spring is- inside the elastic limit, the force required to produce a given displacement is proportional to the displacement. If it 32 EQUATIONS WITH X WANTING. requires a fore, P, to produce a displacement of one unit, the force producing a displacement y is Py. If a body attached to a spring is displaced a distance y, the spring exerts on this body a force — Py, the negative sign being used because the force is in a direc- tion opposite to the displacement. Since the product of the mass of a body, multiplied by its acceleration in any direction, measures the resultant force acting on the body in that direction we have M^ = ^Py, (1) In order that equation (1) may hold it is necessary of course that the units of mass and force be such that unit force produces unit acceleration in unit mass. If the mass is in grams, the force must be in dynes. If the mass is in pounds, the force must be in pound- als. If it is desired to use the pound force as the unit, the corre- sponding mass is the Engineer's Unit of 32.2 pounds, for which the name of gee-pound has been suggested by Mauer. ds Calling -— V, and integrating as shown in Art. 26, we get 2 2 To obtain the constant Ci we will call maximum displacement a. Consequently when 31 = o, w is zero and Ci =^ . Hence we have ^ = ^(a^_yO (3) It may be easily shown that (3) is an energy equation. — s~^* P-^ the kinetic energy of the movmg mass. — g— is the work done m displacing it a distance y, and -— - is the work in displacing to a distance, a, from the origin. Transposing the second term, we get 2 2 2 ' which states that the sum of the kinetic and potential energies is a constant, and is equal to the work done in producing the maximum deflection. In fact, we might have begun by writing this equation from the principle of conservation of energy. Changing v to -^ and solving equation (3) we get dt If we begin to count time from a position when y = 0, we have DIFFERENTIAL EQUATIONS. 33 sin— 1 -Z.= and, consequently, under these conditions C^ = 0, and .= V:^sin-i^. ,5, Vllf - -t is equal to 0, t, 2t 3t, etc., y ^ 0, hence r' ' = ''V^. (^) gives the time of a single vibration, that is the time required for the mass to move from the middle of its path to one extremity and back again to the middle. The time of a complete period is -V-^- Instead of counting time from the position where y = 0, we might use ji ^ o as the initial position. In equation (4) Cz becomes 1 1. ^___sm- Hence . = ^4:(sin-t-f-^)=V=?cos-l^- («) J/ = o cos.^-^. (9) EXAMPLES. 1. Prove, by differentiation, that equations (8) and (9) are solutions of equation (1). 2. What is the time of a complete vibration of a 2-pound mass attached to a helical spring, if a force of 32 poundals stretches the spring one foot? Ans. 1.57 sec. 3 D B CHAPTER IV LINEAR DIFFERENTIAL EQUATIONS OF HIGHER ORDERS WITH CONSTANT COEFFICIENTS. 28. A linear differential equation of the »'* order with constant coefficients is of the form ^+^^£^-^^£^+- • +^-=/<-).(^) Ai, A2, . . . An being constants. THE OPERATOR D. 29. In problems of this sort, it is customary to represent the operation of differentiating and dividing by the increment of the in- dependent variable, by the letter D. That is, when .*■ is the independ- ent variable,—; — is equivalent to D, , . ■ is equivalent to If etc. ax djr The letter D when used in this way, is called the symbolic operator. Using the symbolic notation, equation (1) becomes (/)» +^iZ?»-i + ^jZ?».-2+ . . +A„)y=:/{x). (2) An equation such as ,^. + 3-^ -\-2y ^x', ax' ax is represented in this notation by {D-' + SD + 2)y= x'. Expressions w.ritten in the symbolic notation may be factored like ordinary algebraic quantities. Take first the special case (£»^ + 3£> + 2)3,. If D were an algebraic quantity, the factors of this would be (i? + 2)(£) + l)3,. We will show that, when D is the symbolic operator, (D + 2) (£> + 1)3- = {D' + 3C + 2)3-. (D + l)ji is equivalent to-^ -\- y. The expression, (D + a). pre- fixed to any quantity means that the operation — ^ is to be performed ax on the quantity and a times the quantity added to the derivative. If, in this way, we perform the operations indicated by {D + 2) on ( -^ + y) we get which proves the proposition for this special case. 34 DIFFERENTIAL EQUATIONS. 35 EXAMPLES. Prove by performing the operations 1. (D-2)(D + 2)3)=(Z)^-4)3i, 2. (£)— a)(r> — &)3'=(0'— (o + fc)0 + a&)ji. 3. (D-a)(D-&)(i?-c)y=^-(« + * + ^)-0- + {ah + Of + bc)^ — abcy. Example 3, above, ic a general proof of the proposition for expressions of the third order. The terms a, b, and c are the roots of the general differential equation of the third order of the form of equation (2), regarding D as an algebraic quality, and omitting fix). SOLUTION OF THE DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS. 30. A linear differential equation of the second order with constant coefficients may be written {D-a){D-b)y = fM, '(1) where o and 6 are' the roots of D'' + A,D + A, = 0, regarding D as an algebraic quanttiy. In the first member of equation (1), let us represent (D — b)y by a new variable s. The equation then becomes (D-a)2=f(x) (2) This is equivalent to the linear the solution of which is z = e"^ \e—a'^/(x)dx + C^ea^ (3) Now substitute for z its value, {D — b)y, and get the linear r^—by — e<^x \e-axf(x) dx + C.^"^ dx J From this we get y z=. ei^(e('^-i>)xC e-axf(x)dx^ + Cj«o^ + C^eix , (4) In a similar way, in an equation of the form of {D-a){D~b){D-c)y = f{x), everything of the first member to the right of (13 — a) is repre- sented by a single variable for the first step. Then, for the second step, (.D — c)y is represented by a second variable. These two 36 EQUATION WITH CONSTANT COEFFICIENT. operations give a result of the form of equation (4). The final solution is y = C^e"^ + CgHix + C^ec^ir + ecxf e{6- c)x Ce[a -b)x Cg- ax/ ( X ) dx^. ( 5 ) • Equations of higher order may be solved in a similar vyay. The last term of equation (5) is called the particular integral while those terms which contain arbitrary constants, are together called the complementary function. If fix) is zero, the particular integral vanishes, and the solution is the complementary function alone. THE COMPLEMENTARY FUNCTION. 31. The method of finding the complementary function is seen at once from equation (5) of Art. 30. In an equation of the ntli order it consists of » terms of the form C^eax -)- C^eix -|- Cgccx -j. etc., where a, b, c, etc., are the roots of the algebraic equation (2) of Art. 29. EXAMPLES. Solve the first two examples by the method of Art. 30 without using equation (5). 1. iD' + 3D + 2)3- = g^, y= C^e-'^ + C^e-^ + -f . To verify the solution, perform the operation (Z?" + 3/) + 2) on the value of y. {iy^ + 3£) + 2)y= {4: — 6 + i)Cie-2x + (1 _ 3 + 2) (Taf-^ + (1 + 3 + 2)^ = e^. •^ 1 ^2 ^ 12 ^ 144 y =. C^t^x + Cj.f2* + «-l + ^2Z?K-2-j. . . . . -f^„=0 HAS EQUAI, ROOTS. 32. If a pair of equal factors occur we get for the particular integral a pair of terms of the form DIFFERENTIAL EQUATIONS. 37 C^eax -\- C^ei'x^ which is equivalent to Ci + C2 together form a single arbitrary constant and we have then from an equation of the «*'' order only (n — 1) constants, whereas the number of arbitrary constants should ^correspond with the order of the equation. We will therefore investigate to see if equation (5) of Art. 30 applies to this case. Take the equation (D — o)'(£i — c)}i = 0. (1) For convenience we will arrange the terms so as to bring the equal factors (£> — o) last. (£» — c)'(£) — a)(£» — a)ji = 0. (2) If we let {D — a)y=^z, equation (2) becomes (£> — f) (D — a)2 = 0. (8) Since equation (3) does not contain equal roots it may be solved by means of (5) of Art. 80 giving z = deox + Cie"^, (4) ^ — ay = C^ec=c J^ C^eax ^ (5) ax The integrating factor of equation (5) is «—«•»', hence ye-» + A^Dn — 'i- + 4gI)n — 2 + . . . + An = 0, HAS A PAIR OP IMAGINARY ROOTS. 33. Imaginary roots occur in pairs in the form a ± ib. The corresponding terms of the complementary function are Ci« CH-'*)^ + C8«(a-'4)^. (1) From analytical trigonometry we have cos.ar = ^ , sin.a: = r- i 2 2t hence cos.!ir + i sin;ir = e'^, cos.itr — i sin.if = e—'^. £"^[{€1 + Cs)cosbx + (Ci—Ci)isitibxl (2) If we let Ci + C2 = C'i and (Ci — C2)i=:C2 we get in place of the complex expression of equation (1), the simple trigonometric form e<'x(C\ cosbx + C'isinbx) (3) in which d and C'2 are the arbitrary constants. In order that d and C'l may be real Ci and C2 must be com- plex, so that Ci + C2 is real and Ci — C2 imaginary. As we do not use the original Ci, Cz, we may drop the primes in (3) and write it eax(Ci cosbx -\- Ci sin&:i:). EXAMPLES. Solve and verify the results. 1. (ly—W + 13)3; = 0, y = e2^(Ca cos8.ir + Cs sinS^). 2. — ^ -{-by — O, y = Ci cos i^~F:tr + C2 sin tfT^- Derive the answer of example 2 from that of example 1, Art. 27. 3. £y— 16y = 0, 3; = C,^ + de-ix + G cos2;r + C, sin2.j;. dx THE PARTICULAR INTEGRAL. 34. While the last term of equation (5) Art. 30 shows the form of the general expression for the particular integral, the operations DIFFERENTIAL EQUATIONS. 39 there indicated are frequently difficult to perform. We will there- fore give some special methods which greatly reduce the labor. THE INVERSE OPERATOR. 35. If we write i?« +/4iD«-i. + /42i^»-2 + ^J)k-3 -)_ . . . . + An as /(£))/ the linear equation with constant coefficients becomes where X is any function of x. We will use X in this connection instead of f{x) to avoid confusion with f(D). The particular in- tegral is then written X f(.D) ■ The expression is called the inverse operator, to distinguish it from the direct operator, f(D). The inverse operator indicates those integrations and multiplications which are represented by the last term of equation (5) Art. 30. For example, ~= j xdx; and — is the particular integral of the equation (D — 0)31 = X. U — o The particular integral of a linear equation with constant coeffi- cients when subjected to the direct operation, f(D), gives, as we know, the function of x of the differential equation. That is: ■ ■ represents the particular integral of the equation f(D)y^X, and lj^=X. (^PERATORS COMMUTATIVE. 36. Direct and inverse operators are commutative. It is imma- terial in what ordfr^the operations are performed. Let (_D — a) be a direct operator, and — ;- an inverse oper- D — b ator. We are to prove that (^-«)]^^ = ^<^ — )^- (1) From (4) Art. 30, we have -=el'^Ce-i^Xdx. (2) Hence, operating with (D — a), {T> — a)^^-^^ {6 — a)e6^j'e-6^Xdx+X. (3) 40 OPERATORS COMMUTATIVE. Operating with -= ^ on {DX — ax), we get JJ — o ^"pZa^= e'"'^e-^'DXdx — aei=^ ^e-bxXdx. (4) Integrating by parts, the first term of the second member Ce^bxDXdx — e-bxx + b(e-bxXdx. ( 5 ) Substituting in (4) ^'^~ 1''^ = X -\- (b — a)eb'^ (e-bxXdx. (5) D — b J which is the same as the second member of (3). As any function of D may be made up of factors of the form of (£) — a), the proof may be extended in this way to functions of any degree. A special case of the above is that in which o = 6. The second member (3) reduces to X, thus agreeing with the last equation of the preceding article. RESOLUTION OF INVERSE OPERATORS INTO PARTIAL FRACTIONS. 37. Inverse operators may be resolved into partial fractions in the same way as algebraic quantities. Consider the expression 1 (D — o)(D — &) which regarded as an algebraic quantity breaks up into -J_/'-J_ - _J_Y a — b\D—a D—b) We are to show that 1 X = 1 / 1 _ 1 W (D — a) (D — b) a — b\D—a D — b) operating on (1) with (D — a) we get D — b a — b\ D—b) Operating on (2) with {D — b), X=-^(^D-b-{D-a)^X^X, an identity which proves the proposition for the one case. EXAMPLES. 1. Show that^^^.X is equivalent to^^^-^ . 2. Show that ^P^^ is equal to ^_^ + ^ _^ ^ • (1) DIFFERENTIAL EQUATIONS. 41 It will help to make our ideas of this subject more concrete, if we perform these operations on some simple functions of x, and verify the identities. Take example 2 above, and let X = ;r. •iPx _ 2 ,j. ■which is the particular integral of the equation (/)= — l)y=2. (2) using equation (5), of Art. 30 to find this particular integral, we get .^-^ = le'^je-^je'^dx^ = -2, ( 3 ) In like manner D—1 ^ D-\- 1 ^ ' l*> which is identical with the result of (3). EXAMPLES. Obtain the particular integral by decomposition of fractions and verify the results. 2. -g- - J' = 2^^, J/ = ( Ci + x)e=^ + C^e-x. THE PARTICULAR INTEGRAL FOR SPECIAL FORMS OF fi^x). 38. It is not always easy to find the particular integral by the methods already given. To some forms of X which fortunately are those which occur most frequently in physical problems, special methods may be applied with advantage. CASE WHEN Z = e «■*■. 39. Since the «'* derivative of e<^x is ok go^r^ f(D)e'^x =f(a)e<^x^ (1) f(D)e<'x _ f(a)e<'x „ J(D) f(D) • treating both sides of the equation with the same inverse opera- tor. The first member of (2) is equal to if"^, hence the second member, f(D) and —±—-ei'^ — /(^) /(«) 42 ■ PARTICULAR INTEGRAL FOR SPECIAL FORMS. Take the example the particular integral is e%x _ e%x _ e^x £>^ + 4: ~ 9 + 4 ~ 13 ' and the complete solution is > = Ci cos2j|r + '^2 sin2;ir + ■■ If it happens that o is a root of f(D) =0, then f(a) =0, and yr — - = CO, and the method fails. The method of treating a prob- lem of this kind is seen from the following example. ■^^^ = (1) (.D — 2)XD' — 1) Since = 2 the term (D — 2) renders /^(o)=0. Accordingly we will split equation (1) up into two terms 1 ^^ (OV (£» — 2) ' Z>2— 1 ' * ' The second term, by the method of this article, gives ' e2x _ e^x /?2— 1 3~" Equation becomes then J__i e2x=^fl. (3) 3 (Z? — 2) 3 We see then that when (i3 — a) is a factor of f(D) it should be taken out, the remaining factors treated by the method of this article, and the result thus obtained subjected to the operation — according to the methods of the preceding articles. D — a EXAMPLES. 1. (D\2D + l)y = e^, y^{Cx + C)e-^+^. 2. ^-y = 2, y = C^ex + C^e-x -2. ax' d^y y = (Cr +-|L)^^ + C^e-x. dx' 4, (£>= — 7i?+ 12)3i = e^. + «3;r + 5, J, = (G - x)fix + c,eix + ^ + ^. 6. iD — iy{D — \)y = e^xj^ex, DIFFERENTIAL EQUATIONS. 43 6. (D^— l)3; = cosh2T, or J/ = C^ex + C^e—^ + -i- cosh 2^. 7. (Z)-^ — 1)1/ = e'>, J' = C^e=^ + C^f-^ - £^ . 8. (C — l)''3i = cos;r, j,= (Ci^+C)e-=^ - 2^ . (Make use of the relation cos;f^ i j. 9. .^ + 4-^ + 3y = cos 2x, ax ax y = C^^x + C^e-^'' + 8sin2;r-cos2;ir ^ CASE WHERE X — AxMX OX COSajJT. 40. £) waax = a cosax D' sinax = — a' sinax Hence f{D^) smax = f( — a") sinax. (1) Operating on equation (1) with we get f(D') sinax __ f{ — g') sina^ Since the first member of (2) is equal to sina;r we have n-^^=s^nax. Dividing (3) by the algebraic quantity f{ — a') we get sino^r sina^y tcdt ~ fi-a^y Similarly cosgjf _ c6sa;y Take as an example (D' — i)y ^sin;ir the particular integral of which is sin^r _ sinjr sin.» D' — 4: - —1 — 4 5 ■ The complete integral is (2) (3) y - Cii?2» + C^e-^ie ■ sinx ~~5 When, as is generally the case, contains factors of the first 44 PARTICULAR INTEGRAL FOR SPECIAL FORMS. degree of the form -=r these may be brought to the second D — X degree Dy operating on numerator and denominator with (J3-\-x). For example sin2;r _ (D + lj Ax&x _ (£'^ + 3)(£»-y (D= + 3) (£»=-!) (D + 1) sin2:i: _ 2 cos2;r + sin2;ir 5 5 If quadratic factors occur in containing both C and D. D^ may be replaced by — a' and combined with the constant term and the resulting linear factor treated as in the example above. For example cos2,r _ cos2x cos2;r _ D' + 2D + 3 — 4 + 2Z) + 3 2£> — 1 (2£' + l) cos2;r _ 4 sin2;r — cos2;r W — 1 17 ■ That this is the correct solution is seen by performing the operation (£'' + 2£' + 3) upon the final result. When contains the factor -^-^—, — ^ the method fails for f{D) D' + a' — 00. To evaluate — — - — ^ we will give to a in sinax — a' + a D- + 0= a small increment h. The expression becomes then sm(ax -\- hx) _ sm(ax -\- hx] _ WT^' — (a' + 2ah + h') + a' ~ sin ax cos hx + cos ax sin hx ,■•, 2ah + h' ' As h approaches zero as a limit, cos hx becomes practically unity, %vahx becomes equal to hx, and A^ is negligible. We get then from (1) _ sina^ _ , ^ cosa^ . (2) 2o/i 2o The complementary function of (jy + 0^)3) ^ sinajr is Cr cosa.jr + Ci sina.r. The second term of this complementary function has the same variable factor as the first term of the particular integral. Since the constant Ci may have any value from + 00 to — 00 they may be combined into a single term. The com- plete integral of {D' -\- c^')y^i,max is then Ci cosa.ar + C2 sina^r - X cosar Za as may easily be proven by performing the operation ijy-\-(i?). DIFFERENTIAL EQUATIONS. 46 In a similar way it may be shown that cosajf ^ X %\nax EXAMPLES. Solve and verify results. co&x 1. -A^ — 9ji = cos.a:, J' = C"i.f 3* + (fa^Sir — . air 10 2. .^ - ^ + 6y = sin;., dx^ ax y = C^2a; + C.e^<" + cos;r + sin^ J' 1 -r 2 -r ^^ 3. 4^ + 431 = sin2^, 3» = G cos2;tr + C2 sin2A: —^.S^l. . 4.' ^ + -^ + 2*1 + 2;. = sin3;r, dsfi dx^ dx y^Cr Z0S.2X + C. sin2;r + C,e-^ + 3cos3x - sinS^ _ 6. {D^ — ID -\-\2)y = e^^ + iirAx, y = C^e^^ + Cse4a! + £!f + sinS^r + 7 cos3^. _ u 150 6. ^^ay = cosbx, y = C^-a^' + & sinfc;^ + a cos&^ af.ar 6 + or 7. Integrate «<"'' cos&.j:rf.r by means of inverse operators. The particular integral of the linear of the first order — i, -|- oji = cos&^r (1) dx is expressed by e— ex j eaai cosbxdx. If we equate this to the expression obtained with the ifiverse operator we get e-a-rC eax cosbxdx = =0?^ . (2) •/ D -{- a Multiplying (2) by eo^ Cea^cosbxdx = ga».£2!&£^ = , fl^ . ^ sin&^ + g cosfc^^ -' D -\-a -\- a^ omitting the constant. 8. Solve example 9 Art. 39. i\ cu^ tu„4. cosha.ar' cosho.*- „„j sinha.ir _ sinho.^;. 9. Show that -^^^-^^-^, and ^^^--^^^ 10. {D'—\)y = cosh2;.r, y = C,e^ + C^e-^ + ££!^ . 46 PARTICULAR INTEGRAL FOR SPECIAL FORMS. CASE WHEN X = X*", WHERE »» IS A POSITIVE INTERGER. 41. To find the value of "^^ , expand , „^ into a series in ascending powers of D, and operate with this series on jrw. Since >M is a positive interger, D'"x-'» is a constant, and all terms beyond this are zero, so it is only necessary to carry the series to D'". For example evaluate — = — 1 — D' — D', etc. D' — l 1— D' — {1 + D' + D')x'^— {x^ + 6 x+0) = — x' — 6x. g + 4^ = ^, PXAMPLES. y = Ci COS 2;!r + G sin 2.J: + -L^ x' — ^x' + 1^^ 2. iD' + 3D + 2)y^2 + x\ y = C^e-^ + C,e-2^ + -!-(' x'-ax+ ^\ 3. ^+ay = x*, ax y = Ce-<^=c+ ^ _ 4^^ , \1x^ _ 24^ 24 a% ' n^ /7/t ' /7.5 a a^ a" a* a" 4. Integrate e'"^x™dx by means of inverse operators. As in ex. 7, Art. 40, we find that the integral is jr?" _ -o;r/£!l mx>"—l m(m — I )x>"—'i \ D + a \~fl~ a2 a3 y ' 5. (D^ + D)3- = x', 3; = Ci^-^ -\rC~.+ ^{x* — ^x^-\-12x^ — Ux + 2i\ . D'-+D na + o) £>\ ^ ^ -r ■ ) We may complete the solution in several difierent ways. We might perform the operation — — first and then the operation (1 — D + Z»=— . . ). —^ = —, and n 4 ' A — D + £>= — £>= + £)* . . \^ = ^(^* — 4^^ + lix' — 2ix + 2i\. DIFFERENTIAL EQUATIONS. 47 Again we might combine the factors into ( -— - — l+O — D'^-\-iy . . ) and operate with this on x'. Finally we might perform the opera- tion (^1 — D -\- D' — D" -\- . . ) first and then integrate the re- sult. CASE WHERE X ^^ e"'^ V , V BEING ANY FUNCTION OF X. 42. Take first the case where f{D) is of the first degree. ^ - is the particular 'ntegral of the equation D—a {D — a)y = e'"^K (1) Assume that y^e'"^3, and equation (1) then becomes (D — a)e»'^z = e>"^V. (2) Performing the operation (Z? — a) on e'"=^z in the first number of (2), emx(^B — a-{-m)s = e'>'^V. (3) Dividing by e""^, and performing the operation^- D — a-\- m (4) D — a -\- m We may now multiply by the factor e'"^ and substitute for e'"^z its value 3;. This gives y = e'nx— ^ (5) D — a-\-ni When f(.D) is of the second degree, we have the form (D — a) {D — b)y = e»'^V. (6) We may regard (D — b)y as a single variable u, and write (£) — a)M = e«^F. (7) Applying the results of equation (5) to this u^e'"^— — ^^^— = (D-b)y. (8) D — a-\- m Regarding -= 1 — as a single function of ^ and applying equa- ls — a-\- m tion (5) again we get y=Leni'x : L_ eg) (D — &-h »«)(£» — o-fw) , . ^ ' We proceed in a similar way when f(D) is of higher degree. In any case we find gmx 1/ J/ iT = ^""^ Wt^TT- (10) f(D) /(D-\.m) EXAMPLES. 1. (D^-ir2D + \.)y= fS^cos x. y={CiX+ C^) e-x -^ ^2l( 4 cos ;i: -I- 3 sin ;t: ) 18 EXAMPLES AND PROBLEMS. 2. ( Z?2 — 4D + 3)31 = e2;rcos X, y = C.c^^+C.e--.ff^H. J^_y^e-:^^^ y = C.e-+ C.e— - -^(^' + 2^= + 3x^ + 3.r + A) . Integrate e^^ cos Sjrrfjr by means of inverse operators. ^5^ 5 cos 3;ir + 8 sin ix 34 J^2;Tj^arjr, £^ A;ir« — 6;ir= + 6a- — 3 ) . h X, y — Cex + —( x^ex — Ixe—x — «— ;ir \ . 11 = 8. (Z?2 — 4)ji = ;r^ cos ;t:, 6. -^ — y = X cosh jr, ax 7. -ri = r'' sin jf, ji = C — jr^ cos x -\-'i,x A-a x -\-1 cos ;f. ax y = Ci«2^ + Cse-2^ — ^ cos ;!: + — — X sin ;r — r^cos x. 25 125 PROBLEMS. 1. Solve the problem, Art. 27, by the symbolic method. From the equation M^^ — Pywe get ar y = C^ cos yj-j^^ + G sinyj—t, (1) When f =0, y = 0, /i^«ce Ci = 0, and J! = C2 sin -y-^'^. (2) When sin -J-^r^ "s unity the value of y, which is then the max- imum, is equal to C2. Writing a for this maximum y we have y=asin^^i, (3) 2. Euler's formula for the strength of a column. Fig. 5 represents a column or compression member (drawn hor- izontal for cenvenience) subjected to a compression at the ends. The moment at any section distant x from the left end is — Py, where P is the pressure at each end, and y is the displacement of DIFFERENTIAL EQUATIONS. 4d Figure 5. dy the neutral axis of the section. When y is positive -^ decreases ax as X increases, and — is negative ; and when the deflection is down- dx ward making y negative, -^ increases as x increases, and ■— is. ax ''■" positive. Hence in either case dx ElA2^= -Py. dx^ (1) y = Ci cos -y-gj^ + C2 sin S^-^f^, (2)' Putting in the limiting conditions wc find that 3; = A sin -J-^/^ 1 where A is the maximum deflection, and that-W-^rv^ = »"■, m being an integer. From this we get Euler's formula, P = — — ^ When the column bends in a single curve, as in the figure, n is unity. Column with eccentric load. When as in Fig. 6, the load is not applied at the center of the FiGURB 6. en J; of a column but at a distance k from the center, the differential equation is EI- d'y dx" ■P(k + y), (1> (2). i P 1 P y = Ci cos \i-^^ + Ci sin -\j-^^ — k., ( 3 ) From the condition that y = when x = we find Ci = fe. Hence y = k cos-^-^x + Cisinyj—x — k. (4) 4 D E 60 EXAMPLES AND PROBLEMS. From symmetry it is evident that the maximum deflection is ai the middle ; hence _Z = when x = —- . ax / Differentiating (4) and substituting we get G = fetan^-^J-. Substituting in (4), combining, and transforming, , = fe [sec ^ J^4 cos ^5^. - 4) - l]. The maximum deflection at the middle where x = — - is (6) ^=^^'=V£4-0- (5'«e Merriman, Mechanics of Materials, p. 132.) Beam and one end fixed and with a force applied in any direc- tion to the other end. Fig. 7 shows a beam with the right end fixed horizontally, and Figure 7. with a force P, making an angle a to the left of the vertical, ap- plied at the left end. We take the origin at O, the position of the end of the deflected beam. Resolving the force P into its horizontal and vertical com- ponents and multiplying these by y and x respectively, we get the differential equation EI^ = Py sin a—Px cos a (1) dsr ^n.^_L^-]y=-^xcosa (2) \P sina Ip sina ^ A' EI '^ -\ Ef ^ y = Cxc + CzC -\- X cot o, (3) TJsinff the conditions that y = when x = 0, and ^ = 0, when ° ax X = 1,-me get DIFFERENTIAL EQUATIONS. 51 J- Q cota '4 EI \ EI Substituting in (3) 3; = COto| X (5'^^ Church, Mechanics of Engineering, p. 357). If the direction of the force P is to the right of the vertical malting an angle 0, the moment of the horizontal component is re- versed and the differential equation becomes y = C, cos ^' ^^f X + Gsin ^ ^ ^"^^ x - x cot;6, (2) Putting in the limiting conditions we get : cot/3 r ^ ^^ V^ PsinjS / ■) EI In these problems we have the right end of the beam horizontal. The formula will apply to a beam in any position if the X axis is taken parallel to the fixed end, and the angles a or P be the inclin- ation of the'force to the corresponding Y axis. ELECTRICAL PROBLEMS. Circuits having resistance, inductance and capacity. 45. The E. M. F. equation is the positive direction of the current being towards the condenser. Substituting and multiplying, ^^11+^^1 + ^=^- (2) ' dfi ^ dt Whatever the value of e, the complementary function is the same. The roots of LC£>= + i?C£) + 1 = \^^\r.c, — ^C ^^S R^C^ — ALC 2LC the complementary function is — RC + y RiCi — JLC — RC — \/ R'iC^ — JLC Cje_ + C^e , (3) when ^ R'C — 4LC is real. 52 EXAMPLES AND PROBLEMS. When '^ R'C — ^LC is imaginary, the complementary function is * ^^("^^""^^^ 2LC ' +C3COSV __ ^j. (4, When R'C = iLC, we have the case of equal roots and the com- plementary function becomes (Ci + C^t)e-'2r. (5) The particular integral in any case is ^ G / ft \ LCD' + RCD + l' * ' Discharge of a Condenser. In this case e = and the complementary function is the com- plete integral. We have then only to find the constants for each of the cases above. Case of real, unequal roots : R'C'^iLC. _ RC — ■\rR%Ci — JLC , _ RC + \fRiCi - JLC q = C^e '^^ ^ C,e '^^ (1) We may leave the equation in this form, or we may transform it before determining the constants. In the latter case q = .-^(C'.sinh V^!^^. + C. cosh V^!|^ri^/) (2, where C\ = ^"~^" and C\ = ■ ^' + ^^ ■ From the conditions that when * = 0, g:=Q, and i = we get C. = and C. = ^^=====. g = Qe-^( ^^ =^sinh V^'g'-4^C / * ^ \>JR'C' — iLC 2LC + ''°'^^—2LC / '^' Equation (3) may be reduced in the same way as a similar trigonometric function to ^ ^/R-C — ALC \ 2LC RC / ' To get the current we differentiate one of the expressions for q, preferably (3). The result reduces to DIFFERENTIAL EQUATIONS. 53 — e 2L sinh ^ „ t, ( 5 ) ^IRX' — iLC 'iLC Case of imaginary roots: R'(7 — -^1 sinw* _ _ja_ L CfJ -p=J=^,co.[.,-„.-.(--^)] (4) ,(5) A short time after the circuit is closed the terms .of the comple- mentary functions vanish and (5) gives the value of q. The cur- rent after a short time is expressed by .■ = ^?Tpr3-{"'-"-'[*-cfc]}-'" n !l lllllilllllilliiiil!! Mil ■ I liililiiif^'^ i Ml i lit.'"!''i Til!' ,i.;j, i li! i! II, JIpI ll iiiiiii I PfUiiiiN9--l ilililH^ "■ 1 :i.iMi ■:ir .1 I liilJlH^^ iiiiiii l! m M I i I I i I i llii llllllll