m1208

===================================================================================================Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003.

Tel : 040–27898194/95 Fax : 040–27847334 email : [email protected] website : www.time4education.com Sol.SMM631052/19===================================================================================================

HINTS/SOLUTIONS for M1208(Differential Equations)

Classroom Discussion Exercise

1. (c)

2. (a) r23

2

2

2

dxdy1

dxyd

ie r2

322

2

2

dxdy1

dxyd

order 2 and degree 2

3. (a) y = (A + Bx) e3n ye3x = A + Bx ye3x.3 + e3x.y1 = B(y1 3y) e3x = B(y1 3y) . e3x. 3 + e3x. (y2 3y1) = 0 (y1 3y)( 3) + (y2 3y1) = 0 3y1 + 9y + y2 3y1 = 0 y2 6y1 + 9y = 0

4. (d) y = aex + be2x + ce-3x

y1 = aex + 2be2x – 3ce-3x

= (aex + be2x + ce-3x) + be2x – 4ce-3x

y1 = y + be2x – 4ce-3x

y2 = y1 + 2be2x + 12ce-3x

y2 = y1 + 2 (be2x – 4ce-3x) + 20ce-3x

y2 = y1 + 2 (y1 – y) + 20ce-3x

y2 – 3y1 + 2y = 20ce-3x

(y2 3y1 + 2y) c3x= 20 c(y2 3y1 + 2y) e3x.3 + e3x (y3 3y2 + 2y1) = 0 3y2 9y1 + 6y + y3 3y2 + 2y1 = 0ie y3 7y1 + 6y = 0

5. (b) x2 + (y – a)2 = 1

2x + 2(y – a)y’ = 0 y–a='yx

So the 1st equation gives x2+ 1'y

x2

2

22

2x1

'yx

222 x'yx1

6. (d) The equation of the line 1yx 22 is

y = 2m1mx (1) y1 = m

(1) 211 y1xyy

21

21 y1xyy

7. (c) x 1 y dx y 1 x dy 0

0dy)y1(

ydx)x1(

x

0dy1

y11dx1

x11

log(1 x) x = log(1 y) + y + C (1 x) (1 y) = Ae(x+ y)

8. (d) dy y logxdx

x x

dy logxy

logy x logx 1 C

y A x e

dx

9. (a) 2 dy dyy x y y xdx dx

2ydxdyxy

ydxdyx

yx

dxdxy

dxd

Cyxxy

2kyx

1 y

.

10. (c) 2dyxy x ydx

2dy x x ydx

2dy x x dxy

3 2x xlogy C3 2

11. (d) dxx y 1dy

dy x ydx

Let u x y dy du1dx dx

1 udxdu )u1(

dxdu

dx

u1du

─ log(1-u) = x + C ─ log (1─ x + y) = x + C y = (x-1) + Ce-x.

12. (a) The given equation is the same as

0ydxdyx

2

ydxdyx

xdx

ydy

i.e., ayx .

13. (a) y x dy y x dx

xyxy

dxdy



put y = vx,dxdyxv

dxdy

v + x1v1v

dxdv

x 1v

1v2vdxdv 2

dx1v2v

1vx

dx2

logx = clog1v2vlog21 2

is x2 (v2 2v 1) = cis y2 2xy x2 = c

14. (a)xy

yxdxdy 22

Substitute y = vx

dxdvxv

dxdy

v + xvv1

dxdx 2

xv1

vvv1

dxdv 22

xdxvdv

1

2Cxlog

2v

Cxlog2xy 2

15. (a) y = vx

vtanvdxdvxv

x

dxdvvcot

log sin v = log Cx

.Cxxysin

16. (c)x

yxydxdy 22 =

2

xy1

xy

2v1vdxdvxv

where y = vx

2v1dxdvx

xdx

1v

dv2

Cxlog1vvlog 2

Cx1vv 2

Cx1xy

xy

2

2

222 Cxyxy

222 yxyCx .

17. (b)3yx1yx

dxdy

Put x = X + 1, y = Y + 2

YXYX

dXdY

V1V1

dXdVXV

dXdVX =

v1vv21 2

XdX

1V2VdV1V

2

i.e., CX

dX1V2V

dV1V221

2

CXlog21V2Vlog 2 Y2 + 2XY – X2 = C

i.e., (y – 2)2 + 2 (x – 1) (y – 2) – (x – 1)2 = C x2 – 2xy – y2 + 2x + 6y = k.

18. (a) vvdxdvxv 3 where v =

xy

Clogxlogv21

xdx

vdv

23

Cx = .e2

2

y2x

19. (d) 3dyx y xdx x

x

ydxdyx

2

xxy

dxd

xy =

2x2

+C

20. (b) 2dy y tanx cot x x cot xdx

22 xxtan1ydxdy.xtan

22 xxsecydxdy.xtan

2xxtanydxd

C3xxtany

3

21. (c) xsinxy2

dxdy

I.F =

dx

x2

pdxee

= 2xlog2

x1e .

22. (b) P = 2 tan x, Q = sin x

xseclogxdxtan2Pdx 2

I.F = xsece 2xseclog 2

Solution is y (I.F) = dxF.IQ

y sec2 x = dxxsecxsin 2



= dxxcos

xsin2 = xdxsec.xtan

y sec2x = secx + C

y = cos x + C cos2 x

Given y = 0, when x =3

0 =3

cosC3

cos 2

0 =41C

21

C = 2. y = cos x – 2 cos2 x.

23. (d) axeyax

1dxdy x

P =ax

1

; Q = ex (x + a)

ax

1logaxlogPdx

I.F =

ax1

Solution is

dx

ax1axe

ax1y x

Ceax

y x

y = (x + a) (ex + C).

24. (a) xsinxtany1xtan

y1

dydx

2

cosec x cot x 2y1xeccos

y1

dydx

Put z = cosec x

dydxxcotxeccos

dydz

2y1z

y1

dydz

2y1z

y1

dydz

P =y1 ; Q = 2y

1

ylogPdy I.F =y1

Solution is

dyy1

y1

y1z 2 C

y21

y1z 2

22y = 1 + 2cy2

2y cosecx = 1 + 2cy2

2y = sinx. (1 + 2cy2)

25. (a) 0exdydxy1 ytan2 1

2

ytan

2 y1ex

y11

dydx

1

Here 2

ytan

2 y1eQ

y11P

1

IF = ytan 1e

Solution is

dy.e.

y1ee.x ytan

2

ytanytan 1

11

Put u = ytan 1e

du = dyy1

1.e 2ytan 1

= C2uduu

2

Ce21ex ytan2ytan 11

21a , b = 2

ab = 1

Regular Homework Exercise

1. (c) The given expression can be reduced to3 22 2

2dy d y1 1dx dx

Hence, order = 2 and degree = 2.

2. (a) 22

xdxdy1

.

Degree = 2, order = 1.

3. (c) xx BeAedxdy

yBeAedx

yd xx2

2

0ydx

yd2

2 is the differential equation

4. (b) The general equation is(x – 3) 2 + (y – 3) 2 = a 2

Differentiating with respect to x

2 (x – 3) + 2 (y – 3) 0dxdy

0dxdy

3y3x

5. (c) x2 + y2 = r2 . 2x + 2y y1 = 0

6. (a)

2

2x 1dy

dx y 1

2 2

3 3

3 3

2 2

x 1 dx y 1 dy

x 1 y 1C

3 3

y 1 x 1 C

y x x y xy 3x 3y 3 C



7. (b) sec2 y tan x dy = – sec2 x tan y dx

dxxtanxsecdy

ytanysec 22

log tan y = – log tan x + log C.i.e., tan x tan y = C.

8. (b) dy x tanydx

xdxdy.ycot

C2x)ylog(sin

2

when x = 0, y =2 C = 0

)

2x(1

2

esiny .

9. (c) dy cos x ydx

Substitute x y u

dxdu1

dxdy

ucosdxdu1

ucos1dxdu

=2usin2 2

Cxdu.2ueccos

21 2

x + C2ucot

x + cot C2

yx

10. (a) 2 2dy 1 x y 2xydx

2yx1dxdy

Substitute (x y) = u

dxdu1

dxdy

1 2u1dxdu

2udxdu

Cxudu2

Cxu1

Cxyx

1

11. (b) dy x ydx y x

Substitute, y = vx

v + x vv1

dxdv

xv1

dxdv

xdxvdv

Cxlog2v2

y2 = 2x2 (logx + C) y2 = 2x2(logx + C)

12. (d)y2x3y3x2

dxdy

, put y = vx

u23v32

dxdvxv

;

u23

v2v62dxdVx

2

2vv31dvv23

xdx2

2 0udu

xdx

; u = 1 3v v2

2 logx + log u = logcie x2 u = c x2 (1 3v v2) = Cie x2 3xy y2 = C

13. (d) u = x + 2y

dxdy21

dxdu

=

1u1u21

dxdu =

1u1u3 dxdu

1u31u

Cx1u3log34u

31

i.e., Cx31y6x3log34y2x

Cy2x21y6x3log34

2 log (3x + 6y – 1) = 3 (x – y) + C.

14. (c) x1v

v2dxdv

,

log x =21 v –

21 log v + C

x2 v = CeV. xy = C xy

e

15. (d) x – y = v,dxdv

dxdy1

(v – 1) vdxdv1

v21v1

dv = dx

On integration,

Cxv21log41

2v

or

2yxCxv21log

41 =

2

1Cyx

log (1 – 2v) = 2 (x + y + C1)log (2y – 2x + 1) = 2 (x + y + C1).

16. (d) By definition

17. (b) xsinePdx xsinlogPdx

P = .xcotxsinlogdxd

18. (d) yx2

dxdyx

x2y

dxdyx



2x2

xy

dxdy

I.F = dxx1

e =x1

dxx1.

x2

x1y 2

xy = 1 + cx2

19. (a) The given equation is

nx 1xey1x

ndxdy

P = ,1x

n Q = ex (x + 1)n

I.F = n

dx1x

n

1x1e

dx

1x11xe

x11y n

nxn

= ex + C y = (1 + x)n (ex + C).

20. (a) x5xy

dxdy

I.F =x1

solution is y. dxx1.x5

x1

Cx5yx1

y = 5x2 + cx

(4, 0) lies on this C = 20The curve is y = 5x2 20x(5, a) lies on this gives a = 25

Assignment Exercise

1. (c) Order = 2 and degree = 3

2. (a) y = p + qe6x + re8x

y1 = 6qe6x 8re8x

e8xy1 = 6q e14x 8re8x y2 + y1 e8x.8 = 84 q e14x

(y2 + 8y1) e6x = 84q(y2 + 8y1) e6x. 6+e6x (y3 + 8y2) = 06y2 48y1 + y3 + 8y2 = 0ie y3 + 2y2 48y1 = 0

3. (d) The equation of the circle with centre(0, 0) is x2 + y2 = a2

2x + 2y. 0dxdy

x+yy1 = 0

4. (b)

5. (c) 2 2x ydx xy dy 0 xdx + ydy = 0

C2y

2x 22

x2 + y2 = K

6. (b) 2dy ydx dx

ydy

2

11 x Cy xy 1 ky

7. (d)1yx

1dxdy

, put z = x + y + 1

dxdy1

dxdz

=z

1zz11

dx1z

dzz

Cxdz1z

11z

z – log (z + 1) = x + Cx + y + 1 – log (x + y + 2) = x + C y – log |x + y + 2| = C.

8. (d) Put z = x + y

dxdy1

dxdz

= 1+ zcoszsin

dxzcoszsin1

dz

dx

2zcos

2zsin2

2zcos2

dz2

dxdz

2ztan12

2zsec2

1Cx2ztan1log

1Cx2

yxtan1log

1Cxe2

yxtan1

1Cx ee2

yxtan1

.Ce2

yxtan1 x

9. (a) 0dxyxdyyx

xyxy

dxdy

Put y = vx,dxdvxv

dxdy

x

1v1v

dxdv 2

xdxdv

1v1v

2

xdxdv

1v1dv

1vv

22

cvtanxlog)1vlog(21 12

log x2 (v2+1) + 2 tan1 v = c



log kxytan2yx 122

Given that y = 1 for x = 1 K = log2 +2

log 2log2x

ytan2yx 122

10. (d) Put y = vxv2v1

dxdvxv

2

xv2v1v

v2v1

dxdv 22

.

xdx

v1vdv2

2

– log (1 – v2) = log c x cxv1

12

i.e., cxyx

x22

2

i.e., cxyx 22

i.e., y2 – x2 + c x = 0

11. (c) Homogeneous differential equation.

12. (c) Put y = v x

dxdvxv

dxdy

v + xv1v

dxdv

v1

dxdvx

xdxvdv

clogxlog2v2

)cx(logx2y

2

2

2

2

x2y

e = cx.

13. (d) 2x + y = z

dxdz

dxdy2 z2

dxdz

2zdxdz

dx2z

dz

log (z + 2) = x + C z + 2 = ex eC

2x + y + 2 = kex

y = kex – 2x – 2.

14. (a) It is linear equation

I. F. = pdxe = elog sec x = sec x

y sec x = Cdxxsec2 = tan x + C

y – sin x = C cos x

15. (a) y dx = (y3 x) dy

2yxy1

dydx

, it is a linear Equation is x

P =y1 Q = y2

I. F = pdye =

dyy1

e = elogy = y

xy = cdyyy2

= c4y4 4xy = y4 + k.

Additional Practice Exercise

1. (d) Take log on both side2 log (x +y) = log a + log (x y)

'y1yx

1'y1yx

2

yxy2x2

'y1'y1

yx3

y3x'y1'y1'y1'y1

y’ =yx3xy3

(3x y) y1 = (3y x)

2. (a) Parabola will have equationy2 = 4a (x – b) ; a, b, are constantdifferentiating twice, yy ‘’ + (y‘)2 = 0.

3. (a) log y = loga + bx

6'y.y1

0'y.y

1'.y"y.y1

2

y y” (y’)2 = 0 y .2

2

2

dxdy

dxyd

4. (d) y eax = C1x + C2

y eax.a + eax.y1 = C1ay.eax.a + aeax.y1 + eax.y2 + y1 . eax . a = 0ie y2

2 + 2ay1 + a2y = 0

5. (a) y = ax2 + bx yx1= ax +b y(x2) + (x1 y1) = ay (2x3) + (x2 y1) + x1 y2 + y1 (x2) = 02y xy1 + x2y2 xy1 = 02y = 2xy1 x2 y2

6. (a) Equation of the ellipse is2 2

2 2x y 1 (1)a b

2 22x 2y dy 0

dxa b

22 1a yy

b (2)x

Substituting (2) in (1) we get,

2 2

2 21

x y 1a a yy x

2 21x a y xy



7. (a) xdy ydx xydx

dy dx dxy x

logy + logx = x + C

log (xy) = x + c xy = Aex

8. (c) dy ydx

dy dxdx

logy = x + c y = Aex

when x = 0, y = A = 2 xy 2e

9. (d) xdy ydx dx dy

dxdy1y

dxdyx

1)y(dxd)xy(

dxd

xy y = x+ C xy = x + y + C

10. (c) Substitute u = x-y 2udxdu1

dxdu)u1( 2 dx

u1du

2

'Cxu1u1log

21

x2Aeu1u1

u =1Ae1Ae

x2

x2

x y =

CeCe

x2

x2

11. (c) y f ‘ (x) dx – f (x) dy = y2 dx or

dxy

dy)x(fdx)x('fy2

dxy

)x(fd

cx

y)x(f

f(x) = (x + c) y.

12. (d) y = dx)1x2e( x = ex – x2 + x + C.

13. (d) z = x + y,dxdy1

dxdz

zsecdxdz dxdzzcos

sin z = x + C, ie sin (x + y) = x + C.

14. (a) z = x – y dxdy1

dxdz

z1

dxdz -z dz = dx

0Cx2z2

(x – y)2 + 2x + C = 0.

15. (b) 0y1

dyydxex2

x

2

x

y1

dyy221dxex

16. (a)xy2

dxdy

xdx2

ydy

Solution is xdx2

ydy

log y = 2 log x + log Clog y = log x2 C y = x2 CSince the curve passes through (1, 2)2 = 12 C C = 2The equation is y = 2x2.

17. (a) The given equation is of the form

x11

x1y

dxdy

Here P =x1

1Q,x1

1

. F = 1 + x

y (1 + x) = dxx1.x1

1

= cxdx .

18. (a)y2

1dxdy 2y dy = dx

y2 = x + C ---------- (1)Since (1) passes through (4, 3) C = 5 (1) becomes y2 = x + 5.

19. (c) Put x + y = v,dxdv

dxdy1

11dxdvvcos

vcos1dxdvvcos

dxdvvcos1

vcos

dxdvvcos1

11

dxdv

2vcos2

112

dxdv2vsec

211 2

v tan cx2v

x + y tan cx2

yx

ie y = tan c2

yx

20. (a) y2edxdy e2y dy = dx

Integration gives cx2

e y2



y = 0, x = 529c,c5

21

29ex

y2

when y = 3, x = 9e21 6 .

21. (b) The given equation isy2yx

dxdy

Put y = vx

v + xvx2vxx

dxdy

x

dxdVv2v1

v22

xdxdv

v21v1v2

xdxdv

v11

v211

32

Integrating,

log (1 – 2v) (1 + v)2 = 3xClog

(x – 2y) (x + y)2 = C.

22. (d)x

)yx(dxdy 0

xyx

dxdy

1xy

dxdy

xydxdyx x)xy(

dxd

c2xxy

2

which passes through(2, 1)

4cc22

42

xxy2 8xxy2 2

23. (d) y = vxdxdvxv

dxdy

The differential equation is

3v1

dxdvxv

3

v21dxdvx

.x

dxv21

dv3

24. (a) 0mydxdy

mydxdy mdx

ydy

cmxylog mxeAy .

25. (a) 1xytanxdxdy

Put z = y – x

1dxdy

dxdz

1ztanx1dxdz

ztanxdxdz dxxdzzcot

log sin z + log C =2x2

log C sin z =2x2 C sin z = 2

z2

e

sin (y – x) = 2x2

ke .

26. (d)2dy dy4 4 0

dx dx

2dy 2 0dx

dy 2dx

y 2x C

27. (a) dx = (t2 + t) dt

x = C2t

3t 23

when t = 0, x = 0, C = 0

2t

3tx

23 6x = t2 (2t + 3) (1)

t1dydy

y = log (1 + t)

1 + t = ey t = ey – 1.Substituting in (1), we get6x = (ey – 1)2 [2ey – 2 + 3] 6x = (2ey + 1) (ey – 1)2.

28. (d) The given equation is2SSets2

dsdt is a linear D.E

P = 2s; Q =2sse I.F =

2seSolution is

dseseet222 sss

.C2set

2s2

29. (a) 2y 4ax dy2y 4adx

y2 = xdxdyy2

dyy 2xdx

.

30. (c)2

2dy ydx x 22 x

dxydy

Cx1

y1

(x y) = kxy

;;

m1208

Documents