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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.7 | Set 2

  • Last Updated : 05 Mar, 2021

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. (y2 + 1)dx – (x2 + 1)dy = 0

Solution:

We have,

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

(y2 + 1)dx = (x2 + 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.ey + e-x.ey

 (dy/dx) = ey(ex + e-x)

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

On integrating both sides,

∫e-ydy = ∫exdx + ∫e-xdx

-e-y = ex – e-x + c

e-x-e-y = ex + 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). (xy2 + 2x)(dx) + (x2y + 2y)dy = 0

Solution:

We have,

(xy2 + 2x)(dx) + (x2y + 2y)dy = 0             

x(y2 + 2)(dx) = -y(x2 + 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(y2 + 1) = -log(x2 + 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) + x2y2 = 0

Solution:

We have,

cosecx logy(dy/dx) + x2y2 = 0             

log(y)dy/y2 = -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/y2 = 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)/y2 = 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)2 + c

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


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