DIFFERENTIAL EQUATIONS An equation involving one dependent variable, one or more independent variables and the differential coefficients (derivatives) of dependent variable with respect to independent variables is called a differential equation. ORDER OF A DIFFERENTIAL EQUATION : The order of the highest derivative involved in an ordinary differential equation is called the order of the differential equation. DEGREE OF A DIFFERENTIAL EQUATION The degree of the highest derivative involved in an ordinary differential equation, when the equation has been expressed in the form of a polynomial in the highest derivative by eliminating radicals and fraction powers of the derivatives is called the degree of the differential equation. EXERCISE Il(A) 1 1. Find the order of the family of the differential equation obtained by eliminating the 2 arbitrary constants b and c from xy = cex + be-x + x . Sol. 2 Equation of the curve is xy = cex + be-x + x Number of arbitrary constants in the given curve is 2. Therefore, the order of the corresponding differential equation is 2. Find the order of the differential equation of the family of all circles with their centers 2. 11 at the origin. 2 Given family of curves is x + y = a Diff (1) w.r.t x, 2x+2y.Y1 — (1) , a parameter. Hence required differential equation is x+y.Y1 = 0. Order of the differential equation is 1. 1. Form the differential equation of the following family of curves where parameters are given in brackets. y=c(x—c Diff. w.r.t x,
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2 and c from(l),y= x- ii) xy= aex + xy = aex + be- Diff.w.r.t. x, x —be y + x. Yl = ae diff.w.r.t. x, + y I + XY2 2 -(2) = aex + be = xy .. 2 Yl + xy2 = xy Which is required differential equation. iii) Y = (a + bx) ekx Diff.w.r.t x, Diff.w.r.t. x, kx = kyl + kbe = 2ky —k 2 y which is required differential equation. iv) y=acos(nx+b);(a,b) ans: Y2+n y=0 2. Obtain the differential equation which corresponds to each of the following family of curves. i) The rectangular hyperbolas which have the coordinates axes as asymptotes. Sol. Equation of the rectangular hyperbola is xy=c where c is arbitrary constant. Differentiating w.r.t. x
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dy x—+y=0 ii) The ellipses with centres at the origin and having coordinate axes as axes. Sol. Equation of ellipse is 2 X Diff. 2x a Diff. 2 Y b2 wrtx 2y dy w.r.t. x, b2 b2 2 a Y.YI x Y.Y2 + = 2 a = Y.YI 111. 1. Form the differential equations of the following family of curves where parameters are given in brackets. i) y = ae3x + be4x; (a, b) Sol. y = ae3x + be 4x Differentiating w.r.t x Yl = 3ae3x -k 4be Differentiating w.r.t x, Y2 = 9ae3x + 16be Eliminating a,b from above equations, 16e4x 1 Y2 9 16 — 7 +12 y = 0 which is the required differential equation. ii) y = ax2 + bx ; (a, b) dy 2 — 2x — + 2y Ans: x dx
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iii) ax2 + by 2 = 1 ; (a, b) sol. Given equation is ax 2 + by 1 Differentiating w.r.t. x 2ax + 2bYY1 = 0 Differentiating w.r.t. x 2 (3) — (2) bx(yy'2 + Yl ) -bYYl = O 2 iv) xy=ax +—; (a, b) x d y +2x —2y=0 2 Ans: x 2 dx 2. Obtain the differential equation which corresponds to each of the following family of curves. i) The circles which touch the Y-axis at the origin. Sol. Equation of the given family of circles is x + y + 2gx = 0 , g is arbitrary const .. .(i) x 2 + y 2 = —2gx Differentiating w.r.t. x 2x + 2YY1 Substituting in (i) .. .(ii) x 2 + y 2 = x(2x + 2YY1) by (ii) = 2x2 + 2XYY1 Y Y — 2XYY1 — 2x2 = 0 dy Y 2 —x = 2xy— dx
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ii) The parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis. Sol. Equation of the given family of parabolas is (y — = 4a(x — h)-----(i) where h,k are arbitrary constants Differentiating w.r.t. x 2(y — k)Y1 = 4a (y — k)Y1 = 2a ...(2) Differentiating w.r.t. x (y -k)Y2 +YI -O From (2), y — Substituting in (3) ...(3) Y2 + Yl 2ay2 -I-YI = 0 iii) The parabolas having their foci at the origin and axis along the x-axis. Sol. Given family of parabolas is Y2 = 4a(x + a)-----(i) Diff. w.r.t.x, From (i) and (2), 4—yy' x + 2 dy dy dx dx = 2yy'x+y y
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SOLUTIONS OF DIFFERENTIAL EQUATIONS I.VARIABLES SEPARABLE = f (x, y) . If f(x, y) is a variables separable function, i.e., f(x, y) = Let the given equation be dy = g(x)dx . By integrating g(x)h(y) then the equation can be written as dx h(y) = f (x, y) . This method of finding the solution is known as both sides, we get the solution of variables separable. 1. Find the general solution of I—x2dy+ 1— sol. Given D.E is I—x2dy— 1— y 2 dx Integrating both sides EXERCISE - Il(B) I—x2dy+ Y dx 2 -1 dx 2 I—x sin Y = —sin x + c Solution is sin x + sin y = c, where c is a constant. dy 2y 2. Find the general solution of — dx x dy _ 2y f4Y=2— dx Sol. Il. 1. sol. dx x x Integrating both sides log c + log y = 2 log x 2 log cy = log x Solution is cy = x where c is a constant. Solve the following differential equations. 2 2 2 2
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Integrating both sides 2. 2 tan y dy _ y—x dx dy _ ey 2 = tan- x + tan dy _ dx Sol. dx ex Integrating both sides e c where c is a constant. dx = fe Y dy —e e-Y =e x +c where c is a constant. 3. (ex + l)y dy + (y + l)dx = 0 Sol. (ex + l)y dy = —(y + l)dx 4. ydy dx Integrating both sides 1 dy — y — log(y +1) = log(e-x +1) + log c y — log(y + 1) = log + 1) y = log(y -k 1) + log + 1) y = log c(y + +1) Solution is : e dy x—y dx 2 dy_ x Y + x e Sol. dx Integrating both sides fey •dy= f(ex +x2)dx 3 x Solution is : e 3
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5. tany dx + tanx dy = 0 Sol. tan y dx = —tan x dy dx —dy cos x sin x tan x tan y Taking integration cos y dx = dy sln y cos x dx = sin x cos y — dy sin y 6. Sol. log sin x = —log sin y + log c log sin x + log sin y = log c log(sin x • sin y) = logc sin x • sin y 1 + x dx+ I + y dy=0 1 + dy 1+x dx= Integrating both sides 1 + x dx= x —x 1+x +—sinh 2 x 1+x + y dx 2 2 2 1 = —sinh- x+c 2 y 2 -k log (X-I- 1-kX2 2 dy dx dy dx Sol. y—5y2 = (x +5)— dx x +5 Integrating both sides dy y(1—5y) 1 dy 5 dy Y 1-5y In x +51= In y —In 11—5y I +1nc In I x +5 In 1-5y 1-5y
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2. dy xy + y 8. dx xy + x dy = E—ldx Sol. dx x -1) 1+— dy= 1+— dx x y + log y = x + log x + log c cx Y —x = log — 111. dy 1. dy Sol. 2 (1+x2)xy 2 (1 + x 2 )xy ydy dx 2 2ydy 2xdx 2 x2(1+x2) Integrating both sides 2ydy 1 1 2x dx 2 2 x 1+x log(l + y2) = log x 2 — log(l + x 2) + log c log(l + x 2) + log(l + y 2) = log x 2 + log c Solution is : (1 + + y2) = cx 2 AY+x2 dx Sol. x 2 dx dy dx Integrating both sides — fX2dX
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(I—e 3 Y) X 3 log 3 3 log(l — e-3Y) = x3 + c' (c' = 3c) 3 3y = ex •k (k Solution is: I—e 3. (xy2 + x)dx + (yx2 + y)dy = 0 Sol. (xy2 + x)dx + (yx2 + y)dy = 0 x(y2 + l)dx + y(x2 + l)dy = 0 Dividing with (1 + + y2) xdx ydy 2 Integrating both sides 2 2 1 — [(log(l + x 2) + log(l + y2)l = log c 2 log(l + x 2 + y 2) = 210gc = log c 2 (1 + + y2) = k when k = c . 2 = 2y tanh x 4. dy Sol. = 2y tanhx = 2 tanh xdx Integrating both sides =ccosh x 5. log y In y sin-I = 2 ftanh x dx = 210g Icosh xl + log c = 2 In cosh x + In c y dy dx = sin(x + y) x + y dy dx dx dt dx dt —— sin t — = 1 + sin t dx
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dt = dx 1 + sint Integrating both sides dt 1 + sin t 1 — sin t dt=x+c cos t sec tdt— Itan t • sect dt — —x+c tan t — sec t = x + c tan(x + y) —sec(x + y) — —x+c 6. dx x + x +1 —dy dx Integrating both sides dy dx dy dx 2 2 1 3 4 2 1 3 2 4 I (y+1/2) 2 2 = — tan 1 4-1 | 2 y -1-1 tan + tan 7. {I = tan (x + y) Sol. I-I = tan (x + y) put v = —1+4Y=1+tan v = sec v dx dx dv cos v•dv=x+c sec v (1 + cos 2v) dv=x+c 2 + cos 2v)dv = 2x + 2c 3/2
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sin 2v = 2x + 2c 2 2v+ sin 2v = 4x + c' 2(x + y) + sin 2(x + y) = 4x + c' 1 2 Homogeneous Differential Equations : dy _ f (x, y) A differential equation is said to be a homogeneous differential equation in x, y if dx g(x, y) both f(x, y), g(x, y) are homogeneous functions of same degree in x and y. dy dv To find the solution of the h.d.e put y = vx, then . Substituting these values in dx dx given differential equation, then it reduces to variable separable form. Then we find the solution of the D.E. NOTE: Some times to solve the give homogeneous differential equation, we take the substitution x =VY• 1. Express xdy — ydx = Sol. x •dy—ydx = dy dx x y x EXERCISE - Il(C) dx in the form F x 2 2 x dy dx dy dx dy dx dy _ Y dx x 2 2 x Which is of the form 2. Express x — yTan dx + x tan x x —dy = 0 in the form F x x dy dx • tan Ans: dy _ x dx tan x x
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dy = y(log y — log x + 1) in the form F 3. Express x • dx x dy _ Y Ans: — — log — dx x x Il. Solve the following differential equations. dy _ x — y 1. dx x + y dy _ x —y Sol. dx x + y (1) is a homogeneous D.E. Put y = vx dy dv dx dx dv x —vx — v) dx x +vx x(l + v) Y ) = —210gcx dy dx dv I—v dx 1 + v (1 + v)dv 2 1—2v—v 1 2 2 I—v—v—v 2 1—2v v —— log(l — 2v — v2) = log x + log c 2 1 --10g 1-2.—- 2 2 (x 2 — 2xy log 2 = log cx 1 22 = log(cx) 2 x (x x 2 — 2xy — Y 2 x — 2xy — Y 2 = (cx) c
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2. (x2 + y2)dy = 2xy dx dy 2xy sol. which is a homogeneous D.E. dx x + y Put y = vx dy dv dx dx dv 2x(vx) dx x + v x --1 = cx dv dx 1 + v dv = Let 1+v2 dx A V 2v—v—v B I—v v) + CV(I + v) v = 1 1+1 = C(2) c = 1 v = —1 1+1 = B(—1)(2) 2 = —2B B — dv dv dv = log v — log(l + v) — log(l — v) = log :. log = log x + logc = log cx = cx v = cx(l — v2) Y = cx 1—— Solution is: y = c(x (x
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dy 3. sol. —(x2 + 3y2) (3x2 + y2) —(x2 + 3y2) which is a homogeneous D.E. (3x2 + y2) Put y = vx dy dv dx dv dx dv dx dx —(x2 +3v2x2) 3x + v x 2 1+3v 2 2 2 —3v—v —1—3v2 2 —dx x 2 —x 2 (1 + 3v2) 2 c 3 Multiplying with (v + 1) 2 Equating the coefficients of v Equating the coefficients of V 1 2 4 2 3 1 2 4 2 dv = —log x + log c log(v + 1) + — 4 = log— x
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Solution is: log —+1 + = log — 2x 2x log — log 2x +2xy — 2x = log — 2xy = log log c = —log 4. Y2 dx + (x — xy)dy = 0 Sol. y 2 dx + (x dy 2xy — xy)dy = (xy — x2)dy which is a homogeneous D.E. dx xy — x dy dv Let y = vx dx dx dv dv v dx v—l dx v—l dx v — log v = log x + log k v = log v + log x + log k = log k(vx) = log ky ky = e 2x dx Sol. which is a homogeneous D.E. 2x
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dx 2x dv (1 + v2) dy dv dx dx dv (x + vx) I = log cx — v(l — 3v) 1—3v 2x 1 + v +2v—2v dx dv 2 tan dx — 2 tan v = log x + log c 6. (x2 — y2)dx — xy dy = 0 Ans: x2(x2 — 2y2) = k 7. (x2y — 2xy2)dx = (x3 — 3x2y)dy Sol. (x2y — 2xy2)dx = (x — 3x2y)dy dy x y— 2xy which is a homogeneous D.E. dx x3 —3x y dy dv Put y = vx so that — dx dv x v—2v x dx x —3vx v—2v 1—3v v dv —2v (v —2v2)x3 (1 — 3v)x v-2v 1-3v 1-3v 1—3v —2v +3v 1—3v dx dv = dx — 3 log v = log x + log c
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—Y = 3 log —A — log = log x + log c = log xc —E = log xc + log — = log cx • = log cy cy •e 8. y 2 dx + (x2 — xy + y2)dy = 0 tan-l(y/x) Ans: 9. (y2 — 2xy)dx + (2xy — x2)dy = 0 Ans: 10. dx Sol. dx x x xy(y — x) = c x — which is a homogeneous D.E. x dy dv Put y = vx dx dx dv dv —2v dx dv dy v —2v x A Let —2v v v -2)+Bv 1 = A(v dx B
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1 2 1 v = 2 I = 2B B = 2 1 1 1 dv = 1 — [log v — log(v — 2)] = log x + log c 2 1 — log— = log cx 2 log v v 2 = —logcx = log(cx) 2 1 (y/ x) —2 c2x2 1 1 —(y-2x) y—2x c x Solution is . 11. xdy—ydx = Ans: y + x + y 12. (2x — y)dy = (2y — x)dx 2 c 2 y—2x=c x y=kx y where k = c + y dx 2 = cx Ans; (y — x) = c2 (x + y)3. 13. dx Ans: x2 + 2y2 (c + log y) = 0. 2 14. Solve 2— 2 dx x x Ans: (y — x) = cx y
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= —(v + eV)dy dy dv = log(v+ eV) = —log y + logc v + ev Y dy 2. Solve : x sin— = y sin— x dx Y dy Sol. x sin— — y sin— x dx 111. 1. Solve (1+e Sol. -I-ex/ y 1 dy=o Put y = vx dx dy Put dx dy dv dy dv v + veV + —k dy which is a homogeneous D.E. dv dy v — vev = 0 dy dx sin x sin dy dv
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v dv dx 1 sin v — sin v dv vsin v —1— v sin v sin v dx —sin vdv = — dx x —Sin v • = + cos v = log x + log c = log cx cos(y/ x) cos v cx = e 3. Solve: xdy= y+xcos dx . Ans: tan x x = logx+c. 4. Solve : (x — y log y + y log x)dx + x(log y — log x)dy Ans: = (x — y) log x + y log y = (xdy — ydx)y sin 5. Solve (ydx + xdy)x cos x x Ans: xycos x 6. Find the equation of a curve whose gradient is — dx which passes through the point (1, Tt/4). = — cos , where x > 0, y > 0 and x x Sol. = Y — cos — which is homogeneous differential equation. dx x x Put y = vx dy dv dx dx dv dv = v —cos v -4 dx cos v tan v = —log I x I +c sec v = This curve passes through (1, 71/4) — = c —logl=>c=l tan 4
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Equation of the curve is: tan v = 1 — log I x tan I—log I x I x Equations Reducible to Homogeneous Form -Non Homogeneous Differential Equations dy ax + by + c The differential equation of the form is called non homogeneous dx a'x + b'y+ c differential equation.
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