Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.3 | Set 2

Question 11: Show that y=(c-x)/(1+cx) is the solution of the differential equation.

(1+x²)(dy/dx)+(1+y²)=0

Solution:

We have,

 y=(c-x)/(1+cx)       (i)

Differentiating equation (i)  w.r.t x,

dy/dx=(-1-cx+cx-c²)/(1+cx)²

dy/dx=-(c²+1)/(1+cx²)²

L.H.S,

(1+x²)(dy/dx)+(1+y²)

=(1+x2)[-(c2+1)/(1+cx2)2]+[1+(c-x)²/(1+cx)²]

=

Simplify the above equation,

=0/(1+cx)²

=0

So, (1+x²)(dy/dx)+(1+y²)=0

Question 12: Show that y=e^x(Acosx+Bsinx) is the solution of the differential equation.

d²y/dx²-2(dy/dx)+2y= 0

Solution:

we have,

y=e^x(Acosx+Bsinx)      (i)

Differentiating equation (i)  w.r.t x,

dy/dx=e^x(Acosx+Bsinx)+e^x(-Asinx+Bcosx)     (ii)

dy/dx=e^x[(A+B)cosx-(A-B)sinx]                   (iii)

Again differentiating equation (ii)  w.r.t x,

 d²y/dx² =e^x(Acosx+Bsinx)+e^x(-Asinx+Bcosx)+e^x(-Asinx+Bcosx)+e^x(-Acosx-Bsinx)

d²y/dx²=2e^x[Bcosx-Asinx]          (iv)

d2y/dx2=2e^x[(A+B)cosx-(A-B)sinx] -2e^x(Acosx+Bsinx)

d²y/dx²=2(dy/dx)-2y

d²y/dx²-2(dy/dx)+2y= 0

Question 13: Verify that y=cx+2c² is a solution of the differential equation.

2(dy/dx)²+x(dy/dx)-y=0

Solution:

we have,

y=cx+2c²                (i) 

Differentiating equation (i)  w.r.t x,

dy/dx=c            (ii)

L.H.S,

2(dy/dx)²+x(dy/dx)-y=2(c)² +x(c)-cx+2c²

=0

Question 14: Verify that y=-x-1 is a solution of the differential equation. 

(y-x)dy-(y²-x²)dx=0

Solution:

we have,

y=-x-1   (i)

Differentiating equation (i)  w.r.t x,

dy/dx=-1

L.H.S,

=(y-x)dy-(y²-x²)dx

=(y-x)(dy/dx)-(y²-x²)

=(-x-1-x)(-1)-[(-x-1)²-x²]

=(2x+1)-(x²+2x+1-x²)

=(x²-x²+2x-2x-1+1)

=0

Question 15: Verify that y²=4a(x+a) is a solution of the differential equation. 

y[1-(dy/dx)²]=2x(dy/dx)

Solution:

we have,

y²=4a(x+a)        (i)

Differentiating equation (i)  w.r.t x,

2y(dy/dx)=4a

(dy/dx)=(2a/y)

L.H.S,

=y[1-(dy/dx)²]

=y[1-(2a/y)²] 

=y[1-(4a²/y²)] 

=y[(y²-4a²)/y²]

=(4a(x+a)-4a²)/y

=(4ax+4a²-4a²)/y  

=[2x(2a)]/y

=2x(dy/dx)

=R.H.S

Question 16: Verify that y=ce^{tan-1 x} is a solution of the differential equation. 

(1+x²)(d²y/dx²)+(2x-1)(dy/dx)=0

Solution:

we have,

y=ce^{tan-1 x} (i)

Differentiating equation (i)  w.r.t x,

dy/dx=ce^{tan-1 x} *(1/1+x²)   

(1+x2)(dy/dx)=y     (ii)

Again differentiating equation (ii)  w.r.t x,

2x(dy/dx)+(1+x²)d²y/dx²=dy/dx

(2x-1)(dy/dx)+(1+x²)d²y/dx²=0

Question 17: Verify that y=e^{m cos-1x} is a solution of the differential equation. 

(1-x²)(d²y/dx²)-x(dy/dx)-m²y=0

Solution:

we have,

y=e^{m cos-1 x}(i)

Differentiating equation (i)  w.r.t x,

dy/dx=

dy/dx=       (ii)

Again differentiating equation (ii)  w.r.t x,

(1-x²)d²y/dx²=m²y-xdy/dx

(1-x²)d²y/dx²-m²y-xdy/dx=0

Question 18: Verify that y=log(x+1/√(x²+a²))² is a solution of the differential equation. 

(a²+x²)d²y/dx²+x(dy/dx)=0

Solution:

we have,

y=log(x+1/√(x²+a²))² (i)

Differentiating equation (i)  w.r.t x,

dy/dx=

dy/dx=

dy/dx=

(ii)

Again differentiating equation (ii)  w.r.t x,

(√x²+a²)d²y/dx²+(1/(2√x²+a²))*(2x)*(dy/dx)=0

(a²+x²)d²y/dx²+x(dy/dx)=0

Question 19: Show that the differential equation of which y=2(x²-1)+ce^-x2 is the solution 

dy/dx+2xy=4x³

Solution:

we have,

y=2(x²-1)+ce^-x2(i)

Differentiating equation (i)  w.r.t x,

dy/dx=4x+ce^-x2(-2x)

dy/dx=4x-2cxe^-x2(ii)

L.H.S,

=dy/dx+2xy

=4x-2cxe^-x2 -2x(y=2(x²-1)+ce^-x2

=4x-2cxe^-x2+4x³-4x+2xce^-x2

=0

Question 20: Show that y=e^-x+ax+c is the solution of the differential equation.

e^xd²y/dx²=1

Solution:

We have,

 y=e^-x+ax+c              (i)

Differentiating equation (i)  w.r.t x,

dy/dx=-e^-x+a        (ii)

Again differentiating equation (ii)  w.r.t x,

d²y/dx²=e^-1

(1/e^-1)d²y/dx²=1

e^xd²y/dx²=1

Question 21: For each of the following differential equations verify that the accompanying function is a solution in the mentioned domain.

(i) Function, y=ax, Differential equation, x(dy/dx)=y 

Solution:

We have,

y=ax       (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a        (ii)

From equation (i) a=(y/x)

Putting the value of a in equation (i)

(dy/dx)=a

(dy/dx)=(y/x)

x(dy/dx)=y 

(ii) Function, y=±√(a²-x²), Differential equation: x+y(dy/dx)=0 

Solution:

we have,

y=±√(a²-x²)     (i)

Squaring both sides, we have

y²=(a²-x²)

2y(dy/dx)=-2x

x+y(dy/dx)=0 

(iii) Function, y=a/(x+a), Differential equation, y+x(dy/dx)=y²

Solution:

We have,

 y=a/(x+a)      (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a(-1)/(x+a)²

dy/dx=-a/(x+a)²

L.H.S,

 =y+x(dy/dx)

=a/(x+a)-ax/(x+a)² 

=(-ax+ax+a²)/(x+a)²

=a²/(x+a)²

y²

(iv) Function, y=ax+b+1/2x, Differential equation, x³d²y/dx²=1

Solution:

We have,

y=ax+b+1/2x    (i)

Differentiating equation (i)  w.r.t x,

dy/dx=a+1/(-2x²)

dy/dx=a-1/2x² (ii)

Again differentiating equation (ii)  w.r.t x,

d²y/dx²=0-(-2)/(2x³)

d²y/dx²=1/x³

x³d²y/dx²=1

(v) Function, y=(1/4)*(x±a)², Differential equation, y=(dy/dx)²

Solution:

We have,

y=(1/4)*(x±a)²

Differentiating equation (i)  w.r.t x,

dy/dx=(1/4)*2(x±a)

Squaring both side, we have

(dy/dx)²=(1/4)*(x±a)2

(dy/dx)²=y

Summary

Exercise 22.3 | Set 2 focuses on solving first-order differential equations, particularly those that can be solved using the method of separation of variables or by identifying them as exact differential equations. Key concepts covered include:

• Homogeneous differential equations
• Exact differential equations
• Integrating factors
• Orthogonal trajectories
• Particular solutions with initial conditions