RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter27
Exercise27.7
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

Solve the following differential equations:

Question 1. (x – 1)(dy/dx) = 2xy

Solution:

We have,

(x – 1)(dy/dx) = 2xy     

dy/y = [2x/(x – 1)]dx

On integrating both sides,

∫(dy/y) = ∫[2x + (x – 1)]dx

log(y) = ∫[2 + 2/(x – 1)]dx

log(y) = 2x + 2log(x – 1) + c (Where ‘c’ is integration constant)

Question 2. (x+ 1)dy = xydx

Solution:

We have,

(x+ 1)dy = xydx     

(dy/y) = [x/(x+ 1)]dx

On integrating both sides

∫(dy/y) = ∫[x/(x+ 1)]dx

log(y) = (1/2)∫[2x/(x+ 1)]dx

log(y) = (1/2)log(x+ 1) + c (Where ‘c’ is integration constant)

Question 3. (dy/dx) = (e+ 1)y

Solution:

We have,

(dy/dx) = (e+ 1)y        

(dy/y) = (e+ 1)dx

On integrating both sides

∫(dy/y) = ∫(e+ 1)dx

log(y) = (e+ x) + c (Where ‘c’ is integration constant)

Question 4. (x – 1)(dy/dx) = 2x3y

Solution:

We have,

 (x – 1)(dy/dx) = 2x3y     

(dy/y) = [2x3/(x – 1)]dx

On integrating both sides

∫(dy/y) = ∫[2x3/(x – 1)]dx

∫(dy/y) = 2∫[x+ x + 1 + 1/(x – 1)]dx

log(y) = (2/3)(x3) + x+ 2x + 2log(x – 1) + c (Where ‘c’ is integration constant)

Question 5. xy(y + 1)dy = (x+ 1)dx

Solution:

We have,

xy(y + 1)dy = (x+ 1)dx  

y(y + 1)dy = [(x+ 1)/x]dx

(y+ y)dy = xdx + (dx/x)

On integrating both sides,

∫(y+ y)dy = ∫xdx + (dx/x)

(y3/3) + (y2/2) = (x2/2) + log(x) + c (Where ‘c’ is integration constant)

Question 6. 5(dy/dx) = exy4

Solution:

We have,

5(dy/dx) = exy4          

5(dy/y4) = ex  

On integrating both sides,

5∫(dy/y4) = ∫ex 

-(5/3)(1/y3) = e+ c (Where ‘c’ is integration constant)

Question 7.  xcosydy = (xexlogx + ex)dx

Solution:

We have,

xcosydy = (xexlogx + ex)dx       

cosydy = ex(logx + 1/x)dx

On integrating both sides,

∫cosydy = ∫ex(logx + 1/x)dx

Since, ∫[f(x) + f'(x)]exdx] = exf(x)

siny = exlogx + c (Where ‘c’ is integration constant)

Question 8. (dy/dx) = ex+y + x2ey

Solution:

We have,

(dy/dx) = ex+y + x2ey             

(dy/dx) = exe+ x2ey 

dy = ey(e+ x2)dx

e-ydy = (e+ x2)dx

On integrating both sides,

∫e-ydy = ∫(e+ x2)dx

-e-y = e+ (x3/3) + c (Where ‘c’ is integration constant)

Question 9. x(dy/dx) + y = y2 

Solution:

We have,

x(dy/dx) + y = y2      

x(dy/dx) = y– y

[1/(y– y)]dy = dx/x

On integrating both sides,

∫[1/(y– y)]dy = ∫dx/x

∫[1/(y – 1) – 1/y]dy = ∫(dx/x)

log(y-1) – log(y) = logx + logc

log[(y – 1)/y] = log[xc]

(y – 1)/y = xc

(y-1) = yxc (Where ‘c’ is integration constant)

Question 10. (e+ 1)cosxdx + eysinxdy = 0

Solution:

We have,

(e+ 1)cosxdx + eysinxdy = 0          

(cosx/sinx)dx = -[ey/(e+ 1)]dy

On integrating both sides,

∫(cosx/sinx)dx = -∫[ey/(e+ 1)]dy

log(sinx) = -log(e+ 1) + log(c)

log(sinx) + log(e+ 1) = log(c)

log[sinx(e+ 1)] = log(c)

sinx(e+ 1) = c (Where ‘c’ is integration constant)

Question 11. xcos2ydx = ycos2xdy

Solution:

We have,

xcos2ydx = ycos2xdy      

(x/cos2x)dx = (y/cos2y)dy

xsec2xdx = ysec2ydy

On integrating both sides,

∫xsec2xdx = ∫ysec2ydy

x∫sec^2x - ∫[\frac{d(x)}{dx(x)}∫sec^2xdx]dx = y∫sec^2x-∫[\frac{d(y)}{dy}(y)∫sec^2y]dy

xtanx – ∫tanxdx = ytany – ∫tanydy

xtanx – log(secx) = ytany – log(secy) + c (Where ‘c’ is integration constant)

Question 12. xydy = (y – 1)(x + 1)dx

Solution:

We have,

xydy = (y – 1)(x + 1)dx          

[y/(y – 1)]dy = [(x + 1)/x]dx

On integrating both sides,

∫[y/(y – 1)]dy = ∫[(x + 1)/x]dx

∫[1 + 1/(y – 1)]dy = ∫[(x + 1)/x]dx

y + log(y – 1) = x + log(x) + c

y – x = log(x) – log(y – 1) + c (Where ‘c’ is integration constant)

Question 13. x(dy/dx) + coty = 0

Solution:

We have,

x(dy/dx) + coty = 0          

x(dy/dx) = -coty

dy/coty = -(dx/x)

tanydy = -(dx/x)

On integrating both sides,

∫tanydy = -∫(dx/x)

log(secy) = -log(x) + log(c)

log(secy) + log(x) = log(c)

log(xsecy) = log(c)

x/cosy = c

x = c * cosy (Where ‘c’ is integration constant)

Question 14. (dy/dx) = (xexlogx + ex)/(xcosy)

Solution:

We have,

(dy/dx) = (xexlogx + ex)/(xcosy)         

xcosydy = (xexlogx + ex)dx

cosydy = ex(logx + 1/x)dx

On integrating both sides,

∫cosydy = ∫ex(logx + 1/x)dx

Since, ∫[f(x) + f'(x)]exdx] = exf(x)

siny = exlogx + c (Where ‘c’ is integration constant)

Question 15. (dy/dx) = ex+y + x3ey

Solution:

We have,

(dy/dx) = ex+y + x3ey   

(dy/dx) = exe+ x3ey

dy = ey(e+ x3)dx

e-ydy = (e+ x3)dx

On integrating both sides,

∫e-ydy = ∫(e+ x3)dx

-e-y = e+ (x4/4) + c

e-y + e+ (x4/4) = c  (Where ‘c’ is integration constant)

Question 16. y√(1 + x2) + x√(1 + y2)(dy/dx) = 0

Solution:

We have,

y√(1 + x2) + √(1 + y2)(dy/dx) = 0        

y√(1 + x2)dx = -x√(1 + y2)dy

\frac{\sqrt{1+y^2}}{y}dy=-\frac{\sqrt{1+x^2}}{x}dx
\frac{y\sqrt{1+y^2}}{y^2}dy=-\frac{x\sqrt{1+x^2}}{x^2}dx

On integrating both sides,

∫\frac{y\sqrt{1+y^2}}{y^2}dy=-∫\frac{x\sqrt{1+x^2}}{x^2}dx

Let, 1 + y= z2  

On differentiating both sides

2ydy = 2zdz

ydy = zdz

∫\frac{y\sqrt{1+y^2}}{y^2}dy

∫\frac{zdz*z}{z^2-1}

= ∫[z2/(z– 1)]dz

= ∫[1 + 1/(z– 1)]dz

= z + (1/2)log[(z – 1)/(z + 1)]

On putting the value of z in above equation

\sqrt{1+y^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]

Similarly,

∫\frac{x\sqrt{1+x^2}}{x^2}dx = \sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]

\sqrt{1+y^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]+c

\sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]=c     (Where ‘c’ is integration constant)

Question 17. √(1 + x2)(dy) + √(1 + y2)dx = 0

Solution:

We have,

√(1 + x2)(dy) + √(1 + y2)dx = 0                  

\frac{dy}{\sqrt{(1+y^2)}}=\frac{-dx}{\sqrt{(1+x^2)}}

On integrating both sides,
log[y + √(1 + y2)] = -log[x + √(1 + x2)] + logc

log[y + √(1 + y2)] + log[x + √(1 + x2)] = logc

log([y + √(1 + y2)][x + √(1 + x2)]) = logc

[y + √(1 + y2)][x + √(1 + x2)] = c  (Where ‘c’ is integration constant)

Question 18. \sqrt{(1 + x^2 + y^2 + x^2y^2)} + xy(\frac{dy}{dx}) = 0

Solution:

We have,

\sqrt{(1 + x^2 + y^2 + x^2y^2)} + xy(\frac{dy}{dx}) = 0
\sqrt{[(1 + x^2) + y^2(1+x^2)]}+xy(\frac{dy}{dx})=0
\sqrt{[(1+x^2)(1+y^2)]}+xy(\frac{dy}{dx})=0
\frac{y}{\sqrt{1+y^2}}dy=-\frac{\sqrt{1+x^2}}{x}dx
\frac{y}{\sqrt{1+y^2}}dy=-\frac{x\sqrt{1+x^2}}{x^2}dx

On integrating both sides,

∫\frac{y}{\sqrt{1+y^2}}dy=-∫\frac{x\sqrt{1+x^2}}{x^2}dx

Let, 1 + x= z2  

On differentiating both sides

2xdx = 2zdz

xdx = zdz

-∫\frac{x\sqrt{1+x^2}}{x^2}dy

-∫\frac{zdz*z}{z^2-1}

= -∫[z2/(z– 1)]dz

= -∫[1 + 1/(z– 1)]dz

= -z – (1/2)log[(z – 1)/(z + 1)]

On putting the value of z in above equation

=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]

Let, 1 + y= v2  

On differentiating both sides

2ydy = 2vdv

ydy = vdv

= ∫(vdv/v)

= v

On putting the value of v in above equation

= √(1 + y2)

\sqrt{1+y^2}=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]+c

\sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]=c      (Where ‘c’ is integration constant)

Question 19. (\frac{dy}{dx}) = \frac{e^x(sin^2x + sin2x)}{y(2logy+1)}

Solution:

We have,

(\frac{dy}{dx}) = \frac{e^x(sin^2x + sin2x)}{y(2logy+1)}

y(2logy + 1)dy = ex(sin2x + sin2x)dx

On integrating both sides,

∫y(2logy + 1)dy = ∫ex(sin2x + sin2x)dx

2logy*∫ydy-∫[2\frac{d}{dx}(logy)∫ydy]dy+\frac{y^2}{2}=e^xsin^2x+c

Since, ∫ex(sin2x + sin2x)dx = exsin2

Using property ∫[f(x) + f'(x)]ex = exf(x)

y2log(y) – ∫ydy + y2/2 = exsin2x + c

y2log(y) – y2/2 + y2/2 = exsin2x + c

y2log(y) = exsin2x + c  (Where ‘c’ is integration constant)

Question 20. (dy/dx) = x(2logx + 1)/(siny + ycosy)

Solution:

We have,

(dy/dx) = x(2logx + 1)/(siny + ycosy)        

(siny + ycosy)dy = x(2logx + 1)dx

On integrating both sides,

∫(siny + ycosy)dy = ∫x(2logx + 1)dx

∫sinydy + y∫cosydy – ∫{(dy/dy)∫cosydy}dy = 2logx∫xdx – 2∫{\frac{d(logx)}{dx}∫xdx} + ∫xdx

-cosy + ysiny – ∫sinydy = x2logx – ∫xdx + (x2/2) + c

-cosy + ysiny + cosy = x2logx – (x2/2) + (x2/2) + c

ysiny = x2logx + c (Where ‘c’ is integration constant)

Solve the following differential equations:

Question 21. (1 – x2)dy + xydx = xy2dx

Solution:

We have,

(1 – x2)dy + xydx = xy2dx          

(1 – x2)dy = xy2dx – xydx

(1 – x2)dy = xy(y – 1)dx

\frac{dy}{y(y-1)}=\frac{xdx}{(1-x^2)}

On integrating both sides,

∫[\frac{1}{y-1}-\frac{1}{y}]dy=\frac{1}{2}∫\frac{2x}{1-x^2}dx

log(y – 1) – logy = -(1/2)log(1 – x2) + logc

log(y – 1) – logy + (1/2)log(1 – x2) = logc  (Where ‘c’ is integration constant)

Question 22. tanydx + sec2ytanxdy = 0

Solution:

We have,

tanydx + sec2ytanxdy = 0                

tanydx = -sec2ytanxdy

(sec2y/tany)dy = -dx/tanx

On integrating both sides,

∫(sec2y/tany)dy = -∫cotxdx

Let, tany = z

On differentiating both sides

sec2xdx = dz

∫(dz/z) = -∫cotxdx

log(z) = -log(sinx) + log(c)

On putting the value of z in above equation

log(tany) + log(sinx) = log(c)

log[(sinx)(tany)] = log(c)

sinx.tany = c (Where ‘c’ is integration constant)

Question 23. (1 + x)(1 + y2)dx + (1 + y)(1 + x2)dy = 0

Solution:

We have,

(1 + x)(1 + y2)dx + (1 + y)(1 +x2)dy = 0            

\frac{(1+y)dy}{(1+y^2)}=\frac{-(1+x)dx}{(1+x^2)}
\frac{dy}{(1+y^2)}+\frac{ydy}{(1+y^2)}=\frac{-dx}{(1+x^2)}-\frac{xdx}{(1+x^2)}

On integrating both sides,

tan-1(y) + (1/2)log(1 + y2) = -tan-1(x) – (1/2)log(1 + x2) + c

tan-1(y) + tan-1(x) + (1/2)log[(1 + y2)(1 + x2)] = c (Where ‘c’ is integration constant)

Question 24. tany(dy/dx) = sin(x + y) + sin(x – y)

Solution:

We have,

 tany(dy/dx) = sin(x + y) + sin(x – y)       

tany(dy/dx) = 2sin{(x + y + x – y)/2}cos{(x + y – x + y)/2}

tany(dy/dx) = 2sinxcosy

(tany/cosy)dy = 2sinxdx

On integrating both sides,

∫secytanydy = 2∫sinxdx

secy = -2cosx + c

secy + cosx = c (Where ‘c’ is integration constant)

Question 25. cosxcosy(dy/dx) = -sinxsiny

Solution:

We have,

cosxcosy(dy/dx) = -sinxsiny            

(cosy/siny)dy = -(sinx/cosx)dx

cotydy = -tanxdx

On integrating both sides,

∫cotydy = -∫tanxdx

log(siny) = log(cosx) + logc

log(siny) = log(cosx.c)

siny = c.cosx (Where ‘c’ is integration constant)

Question 26. (dy/dx) + cosxsiny/cosy = 0

Solution:

We have,

(dy/dx) + cosxsiny/cosy = 0           

(dy/dx) = -cosx.tany

dy/tany = -cosxdx

cotydy = -cosxdx

On integrating both sides,

∫cotydy = -∫cosxdx

log(cosy) = -sinx + c

log(cosy) + sinx = c (Where ‘c’ is integration constant)

Question 27. x√(1 – y2)(dx) + y√(1 – x2)dy = 0

Solution:

We have,

x√(1 – y2)(dx) + y√(1 – x2)dy = 0               

x√(1 – y2)(dx) = -y√(1 – x2)dy

\frac{ydy}{\sqrt{(1-y^2)}} = \frac{-xdx}{\sqrt{(1 - x^2)}}

On integrating both sides,

\frac{1}{2}∫\frac{2ydy}{\sqrt{(1-y^2)}}=-\frac{1}{2}∫\frac{xdx}{\sqrt{(1-x^2)}}

√(1 – y2) = -√(1 – x2) + c

√(1 – y2) + √(1 – x2) = c (Where ‘c’ is integration constant)

Question 28. y(1 + ex)dy =(y + 1)exdx

Solution:

We have,

y(1 + ex)dy =(y + 1)exdx                 

\frac{ydy}{(y+1)}=\frac{e^xdx}{(1+e^x)}
1-\frac{1}{(y+1)}=\frac{e^xdx}{(1+e^x)}

On integrating both sides,

∫[1 – 1/(y + 1)]dy = ∫exdx/(1 + ex)

y – log(y + 1) = log(1 + ex) + c (Where ‘c’ is integration constant)

Question 29. (y + xy)dx + (x – xy2)dy = 0

Solution:

We have,

(y + xy)dx + (x – xy2)dy = 0               

y(1 + x)dx = -x(1 – y2)dy

[(1 – y2)/y]dy = -[(1 + x)/x]dx

On integrating both sides,

∫[(1 – y2)/y]dy = -∫[(1 + x)/x]dx

∫(dy/y) – ∫ydy = -∫dx/x – ∫dx

log(y) – (y2/2) = -log(x) – x + c

log(x) + x + log(y) – (y2/2) = c (Where ‘c’ is integration constant)

Question 30. (dy/dx) = 1 – x + y – xy

Solution:

We have,

(dy/dx) = 1 – x + y – xy                    

(dy/dx) = (1 – x) + y(1 – x)

(dy/dx) = (1 – x)(1 – y)

dy/(1 – y) = (1 – x)dx

On integrating both sides,

∫dy/(1 – y) = ∫(1 – x)dx

log(1 – y) = x – (x2/2) + c (Where ‘c’ is integration constant)

Question 31. (y+ 1)dx – (x+ 1)dy = 0

Solution:

We have,

(y+ 1)dx – (x+ 1)dy = 0              

(y+ 1)dx = (x+ 1)dy

\frac{dy}{(y^2+1)}=\frac{dx}{(x^2+1)}

On integrating both sides,

∫\frac{dy}{(y^2+1)}=\frac{∫dx}{(x^2+1)}

tan-1y = tan-1x + c (Where ‘c’ is integration constant)

Question 32. dy + (x + 1)(y + 1)dx = 0

Solution:

We have,

dy + (x + 1)(y + 1)dx = 0               

dy/(y + 1) = -(x + 1)dx

On integrating both sides,

∫dy/(y + 1) = -∫(x + 1)dx

log(y + 1) = -(x2/2) – x + c

log(y + 1) + (x2/2) + x = c (Where ‘c’ is integration constant)

Question 33. (dy/dx) = (1 + x2)(1 + y2)

Solution:

We have,

(dy/dx) = (1 + x2)(1 + y2)             

\frac{dy}{(1+y^2)}=(1+x^2)dx

On integrating both sides,

∫\frac{dy}{(1+y^2)}=∫(1+x^2)dx

tan-1y = x + (x3/3) + c

tan-1y – x – (x3/3) = c (Where ‘c’ is integration constant)

Question 34. (x – 1)(dy/dx) = 2x3y

Solution:

We have,

(x – 1)(dy/dx) = 2x3y      

dy/y = 2x3dx/(x – 1)

On integrating both sides,

∫dy/y = 2∫x3dx/(x – 1)

log(y)=2∫[x^2+x+1+\frac{1}{(x-1)}]

log(y) = (2/3)(x3) + 2(x2/2) + 2x + 2log(x – 1) + log(c)

log(y)=loge^{[\frac{2}{3}x^3+x^2+2x]}+log(x-1)^2+logc

y = c|x – 1|2e[(2/3)x3+x2+2x] (Where ‘c’ is integration constant)

Question 35. (dy/dx) = ex+y + e-x+y 

Solution:

We have,

 (dy/dx) = ex+y + e-x+y             

 (dy/dx) = ex.e+ e-x.ey

 (dy/dx) = ey(e+ e-x)

dy/e= (e+ e-x)dx

On integrating both sides,

∫e-ydy = ∫exdx + ∫e-xdx

-e-y = e– e-x + c

e-x-e-y = e+ c (Where ‘c’ is integration constant)

Question 36. (dy/dx) = (cos2x – sin2x)cos2y

Solution:

We have,

(dy/dx) = (cos2x – sin2x)cos2y        

dy/cos2y = (cos2x – sin2x)dx

sex2ydy = cos2xdx

On integrating both sides,

∫sex2ydy = ∫cos2xdx

tany = (sin2x/2) + c (Where ‘c’ is integration constant)

Question 37(i). (xy+ 2x)(dx) + (x2y + 2y)dy = 0

Solution:

We have,

(xy+ 2x)(dx) + (x2y + 2y)dy = 0             

x(y+ 2)(dx) = -y(x+ 2)dy

\frac{ydy}{(y^2+1)}=\frac{-xdx}{(x^2+1)}

Multiplying both sides by 2,

\frac{2ydy}{(y^2+1)}=\frac{-2xdx}{(x^2+1)}

On integrating both sides,

∫\frac{2ydy}{(y^2+1)}=-∫\frac{2xdx}{(x^2+1)}

log(y+ 1) = -log(x+ 1) + log(c)

(y^2+1)=[\frac{1}{(x^2+1)}]c

(y^2+1)=\frac{c}{(x^2+1)}  (Where ‘c’ is integration constant)

Question 37 (ii). cosecx logy(dy/dx) + x2y= 0

Solution:

We have,

cosecx logy(dy/dx) + x2y= 0             

log(y)dy/y= -x2dx/cosecx

On integrating both sides,

∫[log(y)/y2]dy = -∫x2sinxdx

log(y)∫\frac{dy}{y^2}-∫[\frac{d(logy)}{dy}∫\frac{dy}{y^2}]dy=-[x^2∫sinxdx-∫(\frac{d(x^2)}{dx}∫sinxdx)dx]+c

-log(y)/y + ∫dy/y= x2cosx – 2∫xcosxdx + c

-log(y)/y – 1/y = x2cosx – 2[x∫cosxdx – ∫{dx/dx∫cosxdx}dx] + c

-[{log(y) + 1}/y] = x2cosx – 2(xsinx – ∫sinxdx) + c

x2cosx + [{log(y) + 1}/y] – 2(xsinx + cosx) = c

Question 38 (i). xy(dy/dx) = 1 + x + y + xy

Solution:

We have,

xy(dy/dx) = 1 + x + y + xy                 

xy(dy/dx) = (1 + x) + y(1 + x)

xy(dy/dx) = (1 + x)(1 + y)

ydy/(1 + y) = [(1 + x)/x]dx

On integrating both sides,

∫ydy/(1 + y) = ∫[(1 + x)/x]dx

∫[1 – 1/(1 + y)]dy = ∫(dx/x) + ∫dx

y – log(1 + y) = log(x) + x + log(c)

y = log(x) + log(1 + y) + x + log(c)

y = log[cx(1 + y)] + x (Where ‘c’ is integration constant)

Question 38 (ii). y(1 – x2)(dy/dx) = x(1 + y2)

Solution:

We have,

y(1 – x2)(dy/dx) = x(1 + y2)             

\frac{ydy}{(1+y^2)}=\frac{xdx}{(1-x^2)}

On integrating both sides,

∫\frac{ydy}{(1+y^2)}=∫\frac{xdx}{(1-x^2)}

Multiplying both sides by 2,

∫\frac{2ydy}{(1+y^2)}=∫\frac{2xdx}{(1-x^2)}

log(1 + y2) = -log(1 – x2) + log(c)

log[(1 + y2)(1 – x2)] = logc

(1 + y2)(1 – x2) = c (Where ‘c’ is integration constant)

Question 38 (iii). yex/ydx = (xex/y + y2)dy

Solution:

We have,

yex/ydx = (xex/y + y2)dy         

yex/ydx – xex/ydy = y2dy

ex/y(ydx – xdy)/y= dy

ex/yd(x/y) = dy

On integrating both sides,

∫ex/yd(x/y) = ∫dy

ex/y = y + c (Where ‘c’ is integration constant)

Question 38 (iv). (1 + y2)tan-1xdx + 2y(1 + x2)dy = 0

Solution:

We have,

(1 + y2)tan-1xdx + 2y(1 + x2)dy = 0          -(i)         

\frac{ydy}{(1+y^2)}=-[\frac{tan^{-1}x}{2(1+x^2)}]dx

On integrating both sides,

∫\frac{ydy}{(1+y^2)}=-∫[\frac{tan^{-1}x}{2(1+x^2)}]dx               -(ii)

Let, I = ∫[\frac{tan^{-1}x}{2(1+x^2)}]dx

I=tan^{-1}x∫\frac{1}{2(x^2+1)}dx-∫[\frac{d}{dx}(tan^{-1}x)∫\frac{1}{2(x^2+1)}dx]dx
I=tan^{-1}x(\frac{1}{2}tan^{-1}x)-∫\frac{tan^{-1}x}{2(1+x^2)}dx
I=tan^{-1}x(\frac{1}{2}tan^{-1}x)-I

2I = (1/2)(tan-1x)2

I = (1/4)(tan-1x)2

From equation (ii)

(1/2)log(1 + y2) = -(1/4)(tan-1x)+ c

log(1 + y2) + (1/2)(tan-1x)= c

Question 39. (dy/dx) = ytan2x, y(0) = 2

Solution:

We have,

(dy/dx) = ytan2x        

(dy/y) = tan2xdx

On integrating both sides,

∫(dy/y) = ∫tan2xdx

log(y) = (1/2)log(sec2x) + log(c)

y = c(sec2x)1/2

Put x = 0, y = 2 in above equation

c = 2

y = 2(sec2x)1/2

Solve the differential equations(question 40-48):

Question 40. 2x(dy/dx) = 3y, y(1) = 2

Solution:

We have,

2x(dy/dx) = 3y              

2dy/y = 3dx/x

On integrating both sides,

2∫dy/y = 3∫dx/x

2log(y) = 3log(x) + log(c)

y= x3c

Put x = 1, y = 2 in above equation

c = 4

y= 4x

Question 41. xy(dy/dx) = y + 2, y(2) = 0

Solution:

We have,

xy(dy/dx) = y + 2       

ydy/(y + 2) = dx/x

On integrating both sides,

∫ydy/(y + 2) = ∫(dx/x)

∫1 – \frac{2}{(y+2)} dy = ∫(dx/x)

y – 2log(y + 2) = log(x) + log(c)

Put x = 2, y = 0 in above equation

0 – 2log(2) = log(2) + log(c)

log(c) = -3log(2)

log(c) = log(1/8)

c = (1/8)

y – 2log(y + 2) = log(x/8)

Question 42. (dy/dx) = 2exy3, y(0) = 1/2

Solution:

We have,

(dy/dx) = 2exy3     

dy/y= 2exdx

On integrating both sides,

∫dy/y= 2∫exdx

-(1/2y2) = 2e+ c

Put x = 0, y = (1/2) in above equation

-(4/2) = 2 + c

c = -4

-(1/2y2) = 2e– 4

y2(4e– 8) = -1

y2(8 – 4ex) = 1

Question 43. (dr/dt) = -rt, r(0) = r0

Solution:

We have,

(dr/dt) = -rt       

dr/r = -tdt

On integrating both sides,

∫dr/r = -∫tdt

log(r) = -t2/2 + c

Put t = 0, r =r0 in above equation

c = log(r0)

log(r) = -t2/2 + log(r0)

log(r/r0) = -t2/2

(r/r0) = e^{-\frac{t^2}{2}}

r = r0e^{-\frac{t^2}{2}}

Question 44. (dy/dx) = ysin2x, y(0) = 1

Solution:

We have,

(dy/dx) = ysin2x      

dy/y = sin2xdx

On integrating both sides,

∫(dy/y) = ∫sin2xdx

log(y) = -(1/2)cos2x + c

Put x = 0, y = 1 in above equation

log|1| = -cos0/2 + c

c = (1/2)

log(y) = (1/2) – (cos2x/2)

log(y) = (1 – cos2x)/2

log(y) = 2sin2x/2

log(y) = sin2x

y = e^{sin^2x}

Question 45(i). (dy/dx) = ytanx, y(0) = 1

Solution:

We have,

(dy/dx) = ytanx        

(dy/y) = tanxdx

On integrating both sides

∫(dy/y) = ∫tanxdx

log(y) = log(secx) + c

Put x = 0, y = 1 in above equation

0 = log(1) + c

c = 0

log(y) = log(secx)

y = secx

Question 45(ii). 2x(dy/dx) = 5y, y(1) = 1

Solution:

We have,

2x(dy/dx) = 5y             

(2dy/y) = 5dx/x

On integrating both sides

2∫(dy/y) = 5∫(dx/x)

2log(y) = 5log(x) + c

Put x = 1, y = 1 in above equation

2log(1) = 5log(1) + c

c = 0

2log(y) = 5log(x)

y= x5 

y = |x|(5/2)

Question 45(iii). (dy/dx) = 2e2xy2, y(0) = -1

Solution:

We have,

(dy/dx) = 2e2xy         

(dy/y2) = 2e2xdx

On integrating both sides

∫(dy/y2) = 2∫e2xdx

-(1/y) = 2e2x/2 + c

-(1/y) = e2x + c

Put x = 0, y = -1 in above equation

1 = e+ c

c = 0

-(1/y) = e2x

y = -e-2x

Question 45(iv). cosy(dy/dx) = ex, y(0) = π/2

Solution:

We have,

cosy(dy/dx) = e        

cosydy = exdx

On integrating both sides

∫cosydy = ∫exdx

siny = e+ c

Put x = 1, y = π/2 in above equation

sin(π/2) = e+ c

1 = 1 + c

c = 0

siny = ex

y = sin-1(ex)

Question 45(v). (dy/dx) = 2xy, y(0) = 1

Solution:

We have,

(dy/dx) = 2xy       

dy/y = 2xdx

On integrating both sides

∫(dy/y) = 2∫xdx

log(y) = x+ c

Put x = 0, y = 1 in above equation

log(1) = 0 + c

c = 0

log(y) = x2

Question 45(vi). (dy/dx) = 1 + x+ y+ x2y2, y(0) = 1

Solution:

We have,

(dy/dx) = 1 + x+ y+ x2y2     

(dy/dx) = (1 + x2) + y2(1 + x2)

(dy/dx) = (1 + x2)(1 + y2)

\frac{dy}{(1+y^2)}=(1+x^2)dx

On integrating both sides

∫\frac{dy}{(1+y^2)}=∫(1+x^2)dx

tan-1y = x + (x3/3) + c

Put x = 0, y = 1 in above equation

tan-1(1) = 0 + 0 + c

c = π/4

tan-1y = x + (x3/3) + π/4

Question 45(vii). xy(dy/dx) = (x + 2)(y + 2), y(1) = -1

Solution:

We have,

xy(dy/dx) = (x + 2)(y + 2)        

ydy/(y + 2) = (x + 2)dx/x

On integrating both sides

∫ydy/(y + 2) = ∫(x + 2)dx/x

∫[1 – 2/(y + 2)]dy = ∫dx + 2∫(dx/x)

y – 2log(y + 2) = x + 2log(x) + c

Put x = 1, y = -1 in above equation

-1 – 2log(-1 + 2) = 1 + 2log(1) + c

c = -2

y – 2log(y + 2) = x + 2log(x) – 2

Question 45(viii). (dy/dx) = 1 + x + y+ xy2, y(0) = 0

Solution:

We have,

(dy/dx) = 1 + x + y+ xy   

(dy/dx) = (1 + x) + y2(1 + x)

(dy/dx) = (1 + x)(1 + y2)

On integrating both sides

∫\frac{dy}{(1+y^2)}=∫(1+x)dx

tan-1(y) = x + (x2/2) + c

Put x = 0, y = 0 in above equation

tan-1(0) = 0 + 0 + c

c = 0

tan-1(y) = x + (x2/2)

y = tan(x + x2/2)

Question 45(ix). 2(y + 3) – xy(dy/dx) = 0, y(1) = -2

Solution:

We have,

2(y + 3) – xy(dy/dx) = 0   

xy(dy/dx) = 2(y + 3)

ydy/(y + 3) = 2(dx/x)

On integrating both sides

∫[ydy/(y + 3)] = 2∫(dx/x)

∫[1 – 3/(y + 3)]dy = 2∫(dx/x)

y – 3log(y + 3) = 2log(x) + c

Put x = 1, y = -2 in above equation

-2 – 3log(-2 + 3) = 2log(1) + c

c = -2

y – 3log(y + 3) = 2log(x) – 2

y + 2 = log(x)2log(y + 3)3

e(y+2) = x2(y + 3)3

Question 46. x(dy/dx) + coty = 0, y = π/4 at x = √2

Solution:

We have,

x(dy/dx) + coty = 0                       

x(dy/dx) = -coty

dy/coty = -dx/x

On integrating both sides

∫dy/coty = -∫dx/x

∫tanydy = -∫(dx/x)

log(secy) = -log(x) + c

log(xsecy) = c

Put x = √2, y = π/4 in above equation

log|√2.√2| = c

c = log(2)

log(xsecy) = log(2)

x/cosy = 2

x = 2cosy

Question 47. (1 + x2)(dy/dx) + (1 + y2) = 0, y = 1 at x = 0

Solution:

We have,

(1 + x2)(dy/dx) + (1 + y2) = 0            

(1 + x2)(dy/dx) = -(1 + y2)

\frac{dy}{(1+y^2)}=\frac{-dx}{(1+x^2)}

On integrating both sides

∫\frac{dy}{(1+y^2)}=-∫\frac{dx}{(1+x^2)}

tan-1y = -tan-1x + c

Put x = 0, y = 1 in above equation

tan-1(1) = tan-1(0) + c

c = π/4

tan-1y = π/4 – tan-1x

y = tan(π/4 – tan-1x)

y=\frac{tan\frac{π}{4}-tan(tan^{-1}x)}{1+tan\frac{π}{4}tan(tan^{-1}x)}

y = (1 – x)/(1 + x)

y + yx = 1 – x

x + y = 1 – xy

Question 48. (dy/dx) = 2x(logx + 1)/(siny + ycosy), y = 0 at x = 1  

Solution:

We have,

(dy/dx) = 2x(logx + 1)/(siny + ycosy)                         

(siny + ycosy)dy = 2x(logx + 1)dx

On integrating both sides

∫sinydy + ∫ycosydy = 2∫xlogxdx + 2∫xdx

-cosy + y∫cosydy – ∫[(dy/dy)∫cosydy]dy = 2logx∫xdx – 2∫[(\frac{d(logx)}{dx}  ∫xdx]dx + x+ c

-cosy + ysiny – ∫sinydy = x2logx – ∫xdx + x+ c

-cosy + ysiny + cosy = x2logx – x2/2 + x+ c

Put x = 1, y = 0 in above equation

-1 + 0 + 1 = 0 – (1/2) + 1 + c

c = -(1/2)

ysiny = x2logx + x2/2 – (1/2)

2ysiny = 2x2logx + x– 1

Question 49. Find the particular solution of the differential equation e(dy/dx) = x + 1, given that y(0) = 3 when x = 0.

Solution:

We have,

e(dy/dx) = x + 1               

Taking log both sides,

(dy/dx) = log(x + 1)

dy = log(x + 1)dx

On integrating both sides

∫dy = ∫log(x + 1)dx

y = log(x + 1)∫dx – ∫[\frac{d(log(x+1))}{dx}  ∫dx]dx

y = xlog(x + 1) – ∫xdx/(x + 1)

y = xlog(x + 1) – ∫[1 – 1/(x + 1)]dx

y = xlog(x + 1) – x + log(x + 1) + c

y = (x + 1)log(x + 1) – x + c

Put x = 0, y = 3 in above equation

3 = 0 – 0 + c 

c = 3

y = (x + 1)log(x + 1) – x + 3

Question 50. Find the solution of the differential equation cosydy + cosxsinydx = 0, given that y = π/2 when x = π/2.

Solution:

We have,

cosydy + cosxsinydx = 0         

cosydy = -cosxsinydx

(cosy/siny)dy = -cosxdx

On integrating both sides

∫cotydy = -∫cosxdx

log(siny) = -sinx + c

Put x = π/2, y = π/2 in above equation

log|sinπ/2| = -sin(π/2) + c

0 = -1 + c

c = 1

log(siny) = 1 – sin(x)

log(siny) + sin(x) = 1

Question 51. Find the particular solution of the differential equation (dy/dx) = -4xy2, given that y =1  when x = 0.

Solution:

We have,

(dy/dx) = -4xy          

(dy/y2) + 4xdx = 0

On integrating both sides

∫(dy/y2) + 4∫xdx = 0

-(1/y) + 2x= c

Put x = 0, y = 1 in above equation

-1 + 0 = c

c = -1

-(1/y) + 2x= -1

(1/y) = 2x+ 1

y = 1/(2x+ 1)

Question 52. Find the equation of a curve  passing through the point(0, 0) and whose differential equation is  (dy/dx) = exsinx.

Solution:

We have,

(dy/dx) = exsinx            

dy = exsinxdx

On integrating both sides

∫dy = ∫exsinxdx

Let, I = ∫exsinxdx

I = ex∫sinx – ∫[\frac{de^x}{dx}  ∫sinxdx]dx

I = -excosx + ∫excosxdx

I = -excosx + ex∫cosxdx – ∫[\frac{de^x}{dx}  ∫cosxdx]dx

I = -excosx + exsinx – ∫exsinxdx

I = -excosx + exsinx – I

2I = -excosx + exsinx

I = ex(sinx – cosx)/2

y = ex(sinx – cosx)/2

Question 53. For the differential equation xy(dy/dx) = (x + 2)(y + 2), find the solution curve passing through the point (1, -1).

Solution:

We have,

xy(dy/dx) = (x + 2)(y + 2)                  

ydy/(y + 2) = (x + 2)dx/x

On integrating both sides

∫ydy/(y + 2) = ∫(x + 2)dx/x

∫[1 – 2/(y + 2)]dy = ∫dx + 2∫(dx/x)

y – 2log(y + 2) = x + 2log(x) + c

y – x – c = log(x)+ log(y + 2)2 

y – x – c = log|x2(y + 2)2|

Curve is passing through (1, -1)

-1 – 1 – c = log(1)

c = 2

y – x – 2 = log|x2(y + 2)2|

Question 54. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution:

We have,

Let, v be the volume of the sphere, t be the time, r be the radius of sphere & k is a constant

Volume of sphere is given by v = (4/3)πr3 

According to the question (dv/dt) = k

\frac{d[(4/3)πr^3]}{dt} = k

(4/3)π.3r2(dr/dt) = k

4πr2dr = kdt

On integrating both sides

∫4πr2dr = ∫kdt

4π(r3/3) = kt + c

4πr= 3(kt + c)           -(i)

At t = 0, r = 38

4π(3)= 3(0 + c)

c = 36π

At t = 3, r = 6 in equation (i)

4π(6)= 3(kt + 36π)

864π = 9k + 108π

k = 84π

4πr= 3(84πt + 36π)

r= 63t + 27

r = (63t + 27)1/3

Radius of the balloon after t second is (63t + 27)1/3

Question 55. In a bank principal increases at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log 2 = 0.6931).

Solution:

We have,

Let ‘p’ and ‘t’ be the principal and time respectively.

Principal increases at the rate of r % per year.

dp/dt = (r/100)p

(dp/p) = (r/100)dt

On integrating both sides

∫(dp/p) = (r/100)∫dt

log(p) = (rt/100) + c          -(i)

At t = 0, p = 100

log(100) = 0 + c

c = log(100)          -(ii)

If t = 10, p = 2 × 100 in equation (i)

log(200) = (10r/100) + log(100)

log(200/100) = (10r/100)

log(2) = (r/10)

0.6931 = (r/10)

r = 6.931

Question 56. In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e = 1.648).

Solution:

We have,

Let ‘p’ and ‘t’ be the principal and time respectively.

Principal increases at the rate of 5% per year,

(dp/dt) = (5/100)p         -(i)

(dp/p) = (1/20)dt

On integrating both sides

∫(dp/p) = (1/20)∫dt

log(p) = (t/20) + c          -(ii)

At t = 0, p = 1000

log(1000) = c

log(p) = (t/20) + log(1000)

Putting t = 10 in equation in (i)

log(p/1000) = (10/20)

p = 1000e0.5 

p = 1000 × 1.648

p = 1648

Question 57. In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Solution:

We have,

Let numbers of bacteria at time ‘t’ be ‘x’

The rate of growth of bacteria is proportional to the number present

(dx/dt)∝ x          -(i)

(dx/dt) = kx (where ‘k’ is proportional constant)

(dx/x) = kdt

On integrating both sides

∫(dx/x) = k∫dt

log(x) = kt + c          -(ii)

At t = 0, x = x0(x0 is numbers of bacteria at t = 0)

log(x0) = 0 + c

c = log(x0)

On putting the value of c in equation (ii)

log(x) = kt + log(x0)

log(x/x0) = kt           -(iii)

The number is increased by 10% in 2 hours.

x = x0(1 + 10/100)

(x/x0) = (11/10)

On putting the value of (x/x0) & t = 2 in equation (iii)

2 × k = log(11/10)

k = (1/2)log(11/10)

Therefore, equation (iii) becomes

log(x/x0) = (1/2)log(11/10) × t

t=\frac{2log\frac{x}{x0}}{log\frac{11}{10}}

At time t1 numbers of bacteria becomes 200000 from 100000(i.e, x = 2x0)

t=\frac{2log\frac{2x0}{x0}}{log\frac{11}{10}}

t1=\frac{2log2}{log\frac{11}{10}}

Question 58. If y(x) is a solution of the differential equation  [\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx   , and y(0) = 1, then find the value of y(π/2).

Solution:

We have,

 [\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx                             (i)

dy/(1 + y) = -[(cosx)/(2 + sinx)]dx

On integrating both sides

∫dy/(1 + y) = -∫[(cosx)/(2 + sinx)]dx

log(1 + y) = -log(2 + sinx) + log(c)

log(1 + y) + log(2 + sinx) = log(c)

(1 + y)(2 + sinx) = c

Put at x = 0, y = 1

c = (1 + 1)(2 + 0)

c = 4

(1 + y)(2 + sinx) = 4

(1 + y) = 4/(2 + s inx)

y = 4/(2 + sinx) – 1

We need to find the value of y(π/2)

y = 4/(2 + sinπ/2) – 1

y = (4/3) – 1

y = (1/3)

Solve the following:

Question 1: dy/dx = (x + y + 1)2

Solution:

We have,

dy/dx = (x + y + 1)2

Putting x + y + 1 = v

Therefore, dv/dx – 1 = v2

⇒ dv/dx = v+ 1

⇒ 1/(v+ 1) dv = dx

Integrating both sides, we get

∫ 1/(v+ 1) dv = ∫ dx

Question 2: dy/dx cos (x – y) = 1

Solution: 

We have,

dy/dx cos (x – y) = 1

⇒ dy/dx = 1/cos(x – y)

Putting x – y = v

⇒ 1 – dy/dx = dv/dx

⇒ dy/dx = 1 – dv/dx

Therefore, 1 – dv/dx = 1/cos v

⇒ dv / dx = 1 – 1/cos v

⇒ dv/dx = (cos v – 1)/cos v

⇒ cos v/ (cos v – 1) dv = dx

Integrating both sides, we get

∫cos v/(cos v – 1) dv = ∫dx

⇒ -∫(cos v (1 + cosv)) / (1 – cos2 v) dv =∫ dx

⇒ -∫(cos v (1 + cos v)) / (sin2 v) dv = ∫ dx

⇒ -∫(cot v cosec v + cot2 v) dv = ∫ dx

⇒ -∫ (cot v cosec v + cosec2 v – 1) dv = ∫ dx

⇒ -(-cosec v – cot v – v)= x + C

⇒ cosec ( x – y ) + cot ( x – y ) + x – y = x + C

⇒ cosec ( x – y ) + cot ( x – y ) – y = C

⇒ ((1+cos ( x – y )) / sin ( x – y )) – y = C

⇒ cot (( x – y )/ 2) = y + C

Question 3: dy/dx = ((x – y) + 3)/ (2(x – y) + 5) 

Solution: 

We have,

dy/dx = ((x – y) + 3)/ (2(x – y) + 5) 

Putting x – y = v

⇒ 1 – dy/dx = dv/dx

⇒ dy/dx = 1 – dv/dx

Therefore, 1 – dv/dx = (v + 3)/ (2v + 5)

⇒ dv/dx = 1 – (v + 3)/ (2v + 5)

⇒ dv/dx = (2v + 5 – v – 3)/ 2v + 5

⇒ dv/dx = (v + 2) / (2v + 5)

⇒ (2v + 5)/(v + 2)dv = dx

Integrating both sides, we get

∫(2v + 5)/(v + 2) dv = ∫dx

 ⇒ ∫(2v + 4 + 1)/(v + 2) dv = ∫dx

⇒ ∫((2v + 4)/(v + 2) + 1/(v + 2))dv = ∫dx

⇒ 2∫dv + ∫1/(v + 2)dv = ∫dx

⇒ 2v + log |v + 2| = x + C

⇒ 2(x – y) + log | x – y + 2 | = x + C

Question 4: dy/dx = (x + y)2

Solution: 

We have,

dy/dx = (x + y)2

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx – 1

Therefore, dv/dx – 1 = v2

⇒ dv/dx = v2 + 1

⇒ 1/(v2 + 1) dv = dx

Integrating both sides, we get

∫1/(v2 + 1) dv = ∫dx

⇒ tan-1 v = x + C

⇒ v = tan (x + C)

⇒ x + y = tan (x + C)

Question 5: (x + y)2 dy/dx = 1

Solution: 

We have,

(x + y)2 dy/dx = 1

⇒ dy/dx = 1/( x + y)2

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx – 1

Therefore, dv/dx – 1 = 1/v2

⇒ dv/dx = 1/v2 + 1

⇒ v2/(v2 + 1) dv = dx

Integrating both sides, we get

∫v2/(v2 + 1) dv = ∫dx

⇒ ∫v2 + 1 – 1/(v2 + 1) dv = ∫dx

⇒ ∫(1- 1/(v2 + 1) dv = ∫dx

⇒ v – tan-1 v = x + C

⇒ x + y – tan-1 (x + y) = x + C

⇒ y – tan-1 (x + y) = C

Question 6: cos2 ( x – 2y) = 1 – 2dy/dx

Solution:  

We have,

cos2 ( x – 2y ) = 1 – 2dy/dx 

⇒ 2dy/dx = 1 – cos2 (x – 2y)

Let x – 2y = v

⇒ 1 – 2 dy/dx = dv/dx

⇒ 2 dy/dx = 1 – dv/dx

Therefore, 1 – dv/dx = 1 – cos2 v

⇒ dv/dx = cos2 v

⇒ sec2 v dv = dx

Integrating both sides, we get

∫ sec2 v dv = ∫dx

⇒ tan v = x – C

⇒ tan (x – 2y) = x – C

⇒ x = tan (x – 2y) + C

Question 7: dy/dx = sec(x + y)

Solution:  

We have,

dy/dx = sec(x + y)

⇒ dy/dx = 1/cos ( x + y)

Let x + y = v

⇒ 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx -1

Therefore, dv/dx – 1 = 1/cos v

⇒ dv/dx = (cos v + 1)/ cos v

⇒ cos v/(cos v + 1) dv = dx

Integrating both sides, we get

∫ cos v/(cos v + 1) dv = ∫ dx

⇒ ∫ cos v (1 – cos v)/(1 – cos2 v ) dv = ∫ dx

⇒ ∫ cos v (1 – cos v)/sin2 v  dv = ∫ dx

⇒ ∫ (cos v – cos^2 v)/sin2 v  dv = ∫ dx

⇒ ∫(cot v cosec v – cot2 v) dv = ∫ dx

⇒ ∫(cot v cosec v – cosec2 v + 1) dv = ∫ dx

⇒ – cosec v + cot v + v = x + C

⇒ – cosec (x + y) + cot (x + y) + x + y = x + C

⇒ – cosec (x + y) + cot (x + y) + y = C

⇒ ((-1 + cos (x + y)) / sin (x + y)) + y = C

⇒ – tan ((x + y)/2) + y = C

⇒ y = tan((x + y)/2) + C

Question 8: dy/dx = tan (x + y)

Solution: 

We have,

dy/dx = tan (x + y)

dy/dx = sin (x + y)/cos (x + y)

Let x + y =v

Therefore, 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx – 1

Since, dv/dx -1 = sin v/cos v

⇒ dy/dx = sin v/cos v + 1

⇒ dy/dx = (sin v + cos v)/ cos v

⇒ cos v/(sin v + cos v) dv = dx

Integrating both sides, we get

⇒ ∫ cos v/(sin v + cos v) dv =∫ dx

⇒ 1/2 ∫ {(sin v + cos v) +  (cos v – sin v)}/(sin v + cos v) dv = ∫dx

⇒ 1/2 ∫ dv + 1/2 ∫ (cos v – sin v) / (sin v + cos v) dv = ∫dx

⇒ 1/2 v + 1/2 ∫ (cos v – sin v)/(sin v + cos v) dv = x

Putting sin v + cos v = t

⇒ (cos v – sin v) dv = dt

Therefore, 1/2 v + 1/2 ∫ dt/t = x

⇒ 1/2 v + 1/2 log |t| = x + C

⇒ (x + y) + 1/2 log |sin (x + y) + cos (x + y)| = x + C

⇒ 1/2 (y – x) + 1/2 log |sin (x + y) + cos (x + y)| = C

⇒ (y – x) + log |sin (x + y) + cos (x + y)| = 2C

⇒ y – x + log |sin (x + y) + cos (x + y)| = K                         (where K = 2C)

Question 9: (x + y) (dx – dy) = dx + dy

Solution: 

We have, 

(x + y) (dx – dy) = dx + dy

⇒ x dx + y dx -x dy – y dy = dx + dy

⇒ (x + y -1)dx = ( x + y +1) dy

⇒ dy/dx = (x + y -1)/(x + y + 1)

Let x + y = v

Therefore, 1 + dy/dx =dv/dx

⇒ dy/dx = dv/dx – 1

Therefore,  dv/dx -1 = (v – 1)/(v +1)

⇒ dv/dx = (v – 1)/(v +1) + 1

⇒ dv/dx = (v – 1 + v +1)/(v +1)

⇒ dv/dx = 2v / (v +1)

⇒ (v +1) / 2v dv = dx

Integrating both sides, we get

∫(v + 1)/2v dv = ∫ dx

⇒ 1/2 ∫dv + 1/2 ∫1/v dv = ∫dx

⇒ 1/2 v + 1/2 log |v| = x + C

⇒ 1/2 (x + y) + 1/2 log |x + y| = x + C

⇒ 1/2 (y – x) + 1/2 log |x + y| = C

Question 10: (x + y + 1)dy/dx = 1

Solution: 

We have,

(x + y + 1)dy/dx =1

⇒ dy/dx = 1/(x + y + 1)

Let x + y + 1 = v

Therefore, 1 + dy/dx = dv/dx

⇒ dy/dx = dv/dx – 1

Therefore, dv/dx – 1= 1/v

⇒ dv/dx = 1/v + 1

⇒ v/(v + 1) dv = dx

Integrating both sides, we get

∫ v/(v + 1) dv = ∫ dx

⇒ ∫ (v + 1 – 1)/(v + 1) dv = ∫ dx

⇒ ∫ (1 – 1/(v + 1))dv = ∫ dx

⇒ v – log |v + 1| = x + K

⇒ x + y + 1 – log |x + y+ 1 + 1| = x + K

⇒ y – log |x + y + 2| = K – 1

⇒ y – log |x + y + 2| = C1        ( C1 = K – 1)

⇒ y – C1 = log |x + y + 2|

⇒ ey – C1 = x + y + 2

⇒ e/ eC1 = x + y + 2

⇒ e– C ey = x + y + 2

⇒ C ey  = x + y + 2                  (C = e– C1)

⇒ x =  C ey – y – 2  

Question 11: dy/dx + 1 = ex + y  

Solution: 

We have,

dy/dx + 1 = ex + y                      . . . (1)

Let x + y = t

⇒ 1 + dy/dx = dt/dx

Substituting the value of x + y = t and 1 + dy/dx = dt/dx from (1), we get

dt/dx = et 

⇒ e– t dt = dx

⇒ – e– t = x + C 

⇒ – e– (x + y) = x + C         [Since t = x + y]

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