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

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 + e^x)dy =(y + 1)e^xdx

Solution:

We have,

y(1 + e^x)dy =(y + 1)e^xdx                 

\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 = ∫e^xdx/(1 + e^x)

y - log(y + 1) = log(1 + e^x) + 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) = e^x+y + e^-x+y 

Solution:

We have,

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

 (dy/dx) = e^x.e^y + e^-x.e^y

 (dy/dx) = e^y(e^x + e^-x)

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

On integrating both sides,

∫e^-ydy = ∫e^xdx + ∫e^-xdx

-e^-y = e^x - e^-x + c

e^-x-e^-y = e^x + 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). ye^x/ydx = (xe^x/y + y²)dy

Solution:

We have,

ye^x/ydx = (xe^x/y + y²)dy         

ye^x/ydx - xe^x/ydy = y²dy

e^x/y(ydx - xdy)/y² = dy

e^x/yd(x/y) = dy

On integrating both sides,

∫e^x/yd(x/y) = ∫dy

e^x/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

Summary

Exercise 22.7, Set 2 continues the focus on homogeneous differential equations of the first order. These problems involve equations where the right-hand side can be expressed as a function of y/x. The set includes a mix of general solution problems and particular solution problems with given initial conditions. The complexity ranges from simple rational expressions to more involved quadratic forms. The primary solving technique involves the substitution y = vx to transform the homogeneous equation into a separable one. This exercise reinforces students' ability to recognize homogeneous forms, apply the appropriate substitution, and solve the resulting separable equations.