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Engineering Mathematics

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Published in: Mathematics
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Definitions and Solutions of differential equations!

Hari B / Hyderabad

9 years of teaching experience

Qualification: M.Sc (jntuh campus - 2012)

Teaches: Algebra, Business Mathematics, Mathematics, Statistics, B.Tech Tuition, Polytechnic

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  1. I. 1. DIFFERENTIAL EQUATIONS EXERCISE - Il(D) Solve the following differential equations. dy 12x+5y-9 dx 5x+2y—4 Sol. From given differential equation (5x + 2y — 4)dy = —(12x + 5y — 9)dx (5x + 2y — 4)dy + (12x + 5y — 9)dx = 0 5(x dy+y dx) + 2y dy—4 dy+12x dx — 9 dx = 0 Integrating 5xy + y — 4y + 6x — 9x = c. dy -3x-2y+5 2. dx 2x + 3y+5 Sol. From given differential equation (2x + 3y + 5)dy = (—3x — 2y + 5)dx 2x dy + 3y dy + 5dy = —3x dx —2y dx + 5 dx 2x dy + 3y dy + 5dy + 3x dx —2y dx + 5 dx = 0 Integrating 2xy+—y +—x +5y—5x 4xy+ 3y2 +3x2 —10x +10y Solution is 4xy + 3(x2 + y2) — 10(x — y) = c. dy -3x-2y+5 3. dx 2x +3y —5 3x2 +4xy+3y —IOx —10y Ans: which is the required solution. 4. 2(x —3y +1) = 4x — 2y +1 Ans: 2xy — 3y2 — 2x2 + 2y — x = c.
  2. dy _ x —y +2 5. dx x+y—l Sol. 2xy+y —x —2y dy 2x — y +1 6. dx x+2y—3 Ans: xy+y —x —3x—x=c II. Solve the following differential equations. {=x+y+l (2x + 2y + 3) 1. dy Sol. dx 2x + 2y+3 dy Let v = x + y so that dx dx dx 2v+3 dv = dx 3v+4 2 1 3•dv 2 3 =x+c 6v + log(3v + 4) = 9x + 9c Hence solution is log(3x + 3y + 4) = 3x dy 4x + + 5 2. dx 2x + 3y+4 dy 4x + 6y+5 2(2x + 3y)+5 sol. dx 2x +3y+4 Let v = 2x + 3y dv dy dv 3(2v+5) dx dx dx 2 V -l- 8 6 V + 8V + 23 dv = dx 8v+23
  3. 9 dv 8 8v+23 1 —v + —log(8v+ 23) = x + c 64 8 8v + 9 log(8v + 23) = 64x + 64c 8(2x + 3y) — 64x + 910g(16x + 24y + 23) = c' 2x +3y — 8x + —log(16x + 24y + 23) = c" 3x — 6x + —log(16x + 24y + 23) = c" Hence solution is: y— 2x + —log(16x + 24y + 23) = k 3. (2x + y + l)dx + (4x + 2y — l)dy = 0 = 2, bi = 1, a2 -1) 2v-1 dy sol. dx al Let dv dx 2v-1 3(v 4x + 2y —1 bi dv dy v so that — dx dx v +1 3(v 2v—1 2v—1 2v—1 dv = dx dv = 3dx -1) 1 dv = 3 Jdx 2v + log(v —1) = 3x + c 2v — 3x + log(v —1) = c 2(2x + y) — 3x + log(2x + y —1) = c 4x + 2y — 3x + log(2x + y —1) = c Solution is x + 2y + log(2x + y — 1) = c
  4. 4. Sol. 2dy Let v = x + 2y so that dx dx dx 2v+3 dv = dx 1 1 2 2(4v+5) dv = fdx 2 4v + log(4v + 5) = 8x + 8c 4(x + 2y) —8x + + 2y) + 5] = c' 4x + 8y — 8x + log(4x + 8y + 5) = c' 8y — 4x + log(4x + 8y + 5) = c' which is solution of the given differential equation. 5. (x + y — l)dy = (x + y + l)dx Ans:(x — y) + log (x + y) = c Ill. Solve the following differential equations. dy 3y-7x+7 1. dx 3x —7 y —3 Sol. 3 al al Let x = x + h, y = y + k Where 3k-7h+7=O -7k- Solving these equations, h = 0 and k = 1 dy _ 3y-7x 3 = 0 and dX which is a homogeneous differential equation. dx 3x —7 y dv dy Put y = v x dx dx
  5. dv x(3v — 7) dx x(3— 7 v) dv 3v-7 3v-7- -1) Solution is [y —x + dx 3-7v 7v2 —7 3-7v 3-7v dx 2 3—7 v 3-7v dv = 7v2-7 3 dv— x 7 vdv 1 In x = 3 In E1 ——1n I v 2 —1 | 14 log x—log c 14 V -I-1 2 —7 log I v 2 —11— 141n x —Inc x = 310g = 3 ln(v — 1) — 3 ln(v + 1) — 7 ln(v + 1) — 7 ln(v — l) 14 In x —In c = —101n(v + l) — 41n(v ln(v + +1n(v —l) 2 + In x (V +1)5 • (V—I) 2 • x 7 5 2 x x (y—x)2(y+x)5 dy 6x +5y 2. 2x+18y Ans: (3y — 2x — (x + 2y 10x+8y-12 3. 7x+5y —9 Ans: (x + y — (2x + y — = In c -1)5 - 2) = 343c = c”
  6. 4. (x —y — 2)dx + (x — 2Y — 3)dy 0 dy Sol. Given equation is — dx x —2Y —3 al —1, bl —I, a2 —log 11—2v2 1 al -2 al b2 a2 Let x X + h, y Y + k where —h + k + 2 Solving these equations h l, k —1 dY -X+Y —0 and Oand h-2k-3- dX Therefore which is homogeneous differential equation. dX X-2Y dV Put Y VX so that —z V + X dX dX dV X(-I+V) dX xo-2V) ?-2V dV -1+V x dX ?-2V ((1-2V)dV J 2v2 -? 2 dV ?- 2V2 2 J ?-2V2 log I x — 2 V — -l- 2V2 ?-2V dV J ?-2V2 2 + log C 2v2 -? ?-2V + log c 210g I x —log I I— 2V2 2 2 2
  7. 210g I x +10g 11—2 V 2 | 1 log 1 N15 log I-vé + log c X + YUE + log c x-YUä x-YUä log I X2 -2Y2 logc X +YN6 2 x-YUä •.x2-2Y = c x + YUE Substituting X = x—h = x — 1 , Y = y—k = y + 1 (x — 1) 2 —2(y + 1) 2 x-yU5-l-Uä (x 2 —2y2 —2x —4y —1) = c 5. (x — y)dy = (x + y + 1) dx 1 Ans: logc x + y +x+y+— 2 6. (2x + 3y — 8)dx = (x + y — 3)dy 7. dx 8. dx 2x +3y+4 2x +9y-20 6x +2y —10 Ans: c(x + 2y — 5) = (Y — Linear Equations: 2x)2 = (2x — dy A differential equation of the form —+ Py = Q, where P and Q are functions of x only is called dx a linear differential equation of the first order in y.
  8. Bernoulli's Equation dy An equation of the form — + Py — Qyn , where P and Q are functions of x only, is called a dx Bernoulli ' s equation. EXERCISE - Il(E) I. Find the I.F. of the following differential equations by transforming them into linear form. dy —Y sec 2x dx dy Sol. x— —Y sec 2x dx dx —Ly = 2x sec 2x which is linear in y . dy x fpdx —log x = ——logx=e log(l/x) 1 x e 3 Sol y fpdy I.F. = e 1 —x = 2y which is linear equation in x. 1 —logy log(l/y) _ Il. 1. sol. — + y tan x — dx fp dx Solution of the equation is y.1.F. = = IQ. I.F. dx sec x cos x dx = cos x dx y sec x = 1 1 sin 2x + cos 2x)dx — 2 2 2 Solve the following differential equations. dy — + y tan x — — cos x dx dy — cos x which is linear differential equation in y. Jtanx dx _ log(secx) = sec x
  9. 2. 2y = x + sin x • cos x + c cos x Solution is: 2y = x cos x + sin x • cos x + c • cos x dy —+ y sec x = tan x dx dy Sol. —+ y sec x = tan x which is l.d.e. in y dx sec xdx log(sec x+tan x) = sec x + tan x sol is y.1.F. IQ. I.F. dx y(secx + tan x) = tan x(sec x + tan x)dx — f(secx• tan x + tan x)dx — f(secx • tan x + sec2 x —l)dx Solution is y(sec x + tan x) = sec x + tan x— x + c dy — y tan x = ex sec x which is l.d.e. in y. 3. dx tan x dx log cos x sol. I.F. -e = cos x y.1.F. = y.1.F= IQ. I.F. dx Sol is ycosx= ex sec x cos xdx = eXdx dy 4. x —+ 2y = log x dx 2 x 2 Ans: —log x 4 2 5. (1+x2P+y=e tan- x dx dy 1 Sol. — + 2 dx 1+x fpdx Sol is y IF tan- x which is a linear differential equation in y. 2 2 tan- x .. IQ. I.F. dx
  10. tan- x 2 Consider dx — = fe2tdt = Solution is y tan- x 2y•e tan— x f{dx tan- x 2 "(e dx ...(1) put tan- x 2 tan- x dx = dt dy dx dy Sol. dx I.F. = e Sol is 2y = 2x x 2y = 2x x 2 tan- x tan- x 2 tan- x which is l.d.e. in y. 210g x log x ylF- y x 2 — f2x 4x .. .-y.1.F= IQ. I.F. dx 2x dx = dx=x+c Ans: dy 8. x —+ y = (1+ x)ex dx Ans: y •x = x •ex +c dy dy Sol. Sol is 3x 3x f-}-Edx which is linear differential equation in y. log(1+x3) y.1.F. =y.1.F = IQ. I.F. dx
  11. y(1+x3)= f(1+x2) 3 3 dy —x 10. dx Ans: y=e x + ce x 11. dx dy 1 Sol. + 2 = tan x. tan x which is linear differential equation in y. 2 dx 1 + x dx 2 I-PX sol is y IF tan- x Put t = tan tan- x .. IQ. I.F. dx tan- x dx 2 dx x so that dt = 2 R.H.S. — ft •etdt = t •e t t tan- x tan- x Solution is : y.e (tan- x—l)+c —tan- x y = tan- dy _ + y tan x = Sin x . 12. dx Ans : tan xdx = log secx+c Ill. Solve the following differential equations. dy cosx—+ ysinx = sec x 1. dx dy sol. —+ tan x • y = sec x which is l.d.e in y dx Itan x dx log sec x = sec x sol is y.1.F. IQ. I.F. dx
  12. y • secx= sec xdx = J(l+tan x) sec2 xdx tan x = tan x + 3 2. secx • dy = (y + sin x)dx Sol. dx dy dx Sol is Y + sin x = y cosx + sin x • cos x sec x —y cos x = sin x • cosx which is l..d.e in y Icos xdx — sin x y.1.F. = y.1.F= JQ. I.F. dx —sin x —sin x • sin x • cos x •dx —sin x Consider e • sin x • cos x •dx t = — sin x dt = —cos x dx — sin x •sinx•cosxdx = + et •tdt — sin x (— sin x = t • e —e -1-c=e —sin x — sin x (sin x +1) + c sin x or y = —(sin x +1) +c•e dy 3. x log x • —+ y = 210gx dx Ans: y log x = (log +c dx Sol. (x + y +1) dx dx dx —x = y +1 which is l.d.e in x. dy dy Jpdy J-dy sol is x.1.F.= JQ.1Fdy
  13. = -(?+1)? 5. Solve ?(? —1)??— dx dy 1 Sol. ? ? = ? (? —1)2 which is l.d.e in ? dx ?(? —1) dx fpdx ?(?—1) Sol log— log x—log(x—1) y.I.F. fQ. ? — ?? 2 (? —1) 2 ? I.F. dx ? ?? ? (? dx = — 1)dx ????? solution is dx Ans: ? = +?) ?? ? 5 7. Solve 1—? dx 1—? 2 +?(1—?2) Ans ? = dy — (? — 2)? = ? ? (2? 8. ?(? — 1)— dx Ans: ?(? — 1) = ?2(?2 — ? + ?) 4 ? 4 2 -1) dy 2 ? 9. —(? ? + ??) dx dy 2 ? Sol. —(? ? + ??) dx dx = ??+? ? dy
  14. dx —xy=x y dy Which is Bernoulli's equation Dividing with x , 1 dx 1 x 2 dy x dz Put z = ——so that — dy dz dy 1 dx x 2 dy which is linear differential equation in z fydy y2/2 sol z.1.F= IQ. I.F. dy y2/2 3 y2/2 put = t y dy = dt y2/2 Y y2/2 y2/2 Y -y x -y2/2 2 -y2/2 Hence solution is dy 10. —+ x • sin 2y = x dx Ans : tan y — cos Y
  15. 1 dy 11. y + x— Y dx Sol. 1 dy Y dx 1 dy dx y dx x —1/ y dy dx 1 —+—•x= dy y x 1 — which is differential equation in x f 12dy I.F. = e IQ. I.F. dy sol is x.1.F = x e-l/y= z • eZdz = (z —1) ...U) put — = z —j dy = dz l/y l/y l/y Hence solution is xy = 1 + y + cy e PROBLEMS FOR PRACTICE 1. Find the order and degree of the differential equation dx Ans. Degree = 1, Order = 2 d3y 2. Find the order and degree of dx Ans. Degree = 2, Order = 3 dy dx
  16. 3. 4. d2y x 2 dx dy —+ y = 0 has order 2, degree 1. Prove it. dx 6/5 Find the order and degree of d2y 2 dx 2 dx 6/5 3 dy dx 3 dy dx = 6y Sol. Given equation is: 3 d2y dy i.e. 2 dx dx = (6y)5/6 Order = 2, degree = 1 5. Find the order of the differential equations corresponding to y = c(x — where c is an arbitrary constant. Ans. Order = 1 6. Find the order of the differential equation corresponding to y = Aex + Be3x + Ce5x (A, B, C being parameters) is a solution. Ans. Order = 3 7. Form the differential equation corresponding to y = cx — 2c , where c is a parameter. Ans. y = x dx 2 dy dx 8. Form the differential equation corresponding to y = Acos 3x + B sin 3x where A and B are parameters. d2y Ans. 2 dx 9. Form the differential equation corresponding to the family of circles of radius r given by (x — a2) + (y — = r2, where a and b are parameters. Sol. We have : (x — a2) + (y — Differentiating (1) w.r.to x dy 2(x — a) + 2(y — b) — dx Differentiating (2) w.r.to x ...(2) ...(3) 2 dx 2 dy dx
  17. dy From (2) (x —a) = dx Substituting in (1), we get (y-b)2 dy dx (y-b)2 dy dx d2y From (3) (y-h)— dx dy dx d2y dx Substituting in (4) : i.e. r dx dy dx dy dx d2y dx dy dx 10. Form the differential equation corresponding to the family of circles passing through the origin and having centers on Y-axis. 2 le2xy=0 Ans. (x —y ) 11. Express the following differential equations in the form f(x)dx + g(y)dy dx 1 + x 2 dy ii) Y x dx e x 111) dx
  18. dy 2 3y iv) —+x = x •e dx dy 12. Find the general solution of x + y— dx Ans. x + y =2c = c' dy _ x+y 13. Find the general solution of dx Ans. ex + e dy 14. Solve y x dx Ans. y — a(x + a) a 15. Solve dx Ans. y = c (x — dy 16. Solve dx x(210g x + 1) sin y + y cos y Ans. y sin y = x logx + c 17. Find the equation of the curve, whose slope at any point is y/x and which satisfies the condition y = 1 when x = 3. x—3 Ans. y = e 18. Solve y(l + x)dx + x(l + y)dy = 0 Ans. x + y + log(xy) = c = sin(x + y) + cos(x + y) 19. Solve Ans. x = log 1+ tan 2 2 dy 2 20. Solve (x —y) a Ans. y = —log 2 dx
  19. 21. Solve 1+x 1 + y dx + xydy = 0. Ans. log x —log(l+ 1-l-X2+ 1+y dy x —2y+1 22. Solve dx 2x — 4y Ans. (x — 2y)2 + 2x = c' where c dy 23. Solve dx Ans. x+c=2 y—x +210g( y—x —1) dy 24. Solve —+1 = ex + Y dx (X-ky) Ans. x+e 25. Solve 1 1 dy 26. Solve dx = (3x + y + 4) 2 3x + y +4 =x+c —x tan(y —x) = 1 2 Ans. log I sin(y —x) I 2 27. Show that f(x, y) = 1 + ewy is a homogenous function of x and y. 28. Show that f (x, y) = x x + y —y is a homogenous function of x and y. 29. Show that f(x, y) = x — y log y + y log x is a homogeneous function of x and y. dx x x 30. Express (1+eX/Y)dx +e dy = 0 in the form 31. Express (x x + y dy 32. Express — dx x + ye dy —y + xydy = 0 in the form dx x dx x in the form 2x/Y dy
  20. 33. Solve Ans. xy 2xy y X — 2xy 3 34. Solve (x2 + y2)dx = 2xy dy. Ans. cx(x — y ) = x 35. Solve xy dy—(x + y )dx Ans. y = 3x log cx. 2 dy x + y 36. Solve 2 dx 2x Ans. 2x = (x — y)(log x + c) 37. Solve x sec — (ydx + xdy) = ycsc Sol. x sec x x sec dy dx x — (ydx+ xdy) = ycsc — (xdy— ydx) . x — (xdy — ydx) x x dy dx — ycsc x x • sec x —Y • csc x dy dx + x sec y csc x x + x sec —Y ycsc dy x dx x x sec x x — y csc x This is a homogeneous equation. Put y = vx dy dv dx dx dv v csc v + sec v dx v csc v — sec v 1 sin v 1 sin v cos v 1 cos v v(v cosv+ sin v) v cos v — sin v
  21. dv v(v cos v + sin v) dx v cos v — sin v cos v + sin v — v cos v + sin v) v cos v 2v sin v v cos v — sin v v cos v — sin v v sin v cos v sin v 1 — sin v dx x dx x log sin v — log v = 2 log x + log c sin v log x —sin x sin v 2 = logcx = cx =cx sin = cxy x 38. Give the solution of x sin —dx = ydx —xdy which passes through the point (1, It/4). x Ans. cot = log x +1 x 39. Solve (x —3xy2)dx +(3x y Ans. 2 -1 2 x = cx 2 2 x 40. Transform the following two differential equations into linear form. dy Ans. — + py = Q dx 41. (x +2y3)AY dx dx Ans. —+Px=Q dy Find I.F. of the following two differential equations by transforming them into linear form. dy 42. (cosx)—+ ysinx = tan x dx flog sec x Ans. I.F. = e = sec x
  22. dx f2dy 2 2 log y log y Ans. 1.F.=e 2 dy 44. Solve (1+x — + 2xy — 4x dx 3 3 dy 43. Solve 2x —10y — 2(x — 2 Ans. y(1+x2)= f4x 1 dy 45. Solve x dx Ans. 2 dy x 3 2 (I—x)e 46. Solve sin x • + y = cot x . dx — cot x —cot x Ans. y •e = (cot x + 47. Find the equation of the equation x(x condition that y = 9 when x = 3. Ans. = x + 210g(x — 2) x(x — 2) 48. Solve (1+ y2)dx = (tan- y—x)dy. tan- Y tan- Y (tan y —1) + c . Ans. x •e dy dx = x (x —2) which satisfies the

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