RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations

Here we provide RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations

Question 1: Show that y=be^x+ce^2x is the solution of the differential equation.

d² y/dx²-3(dy/dx)+2y=0

Solution:

y=be^x+ce^2x (i)

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

dy/dx=be^x +2ce^2x

dy/dx=be^x+2ce^2x (ii)

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

d²y/dx² =be^x+4ce^2x (iii)

we have,

d²y/dx² -3(dy/dx)+2y=0 (iv)

Putting the values ofd² y/dx² anddy/dx in equation (iv)

=be^x+4ce^2x-3(be^2x+2ce^2x)+2(be^x+ce^2x)

=3be^x-3be^x+6ce^2x-6ce^2x

=0

So,d²y/dx²-3(dy/dx)+2y=0

Question 2: Verify that y=4sin3x is a solution of the differential equation.

d²y/dx²+9y=0 

Solution:

y=4sin3x (i)

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

dy/dx=(4)(3)cos3x (ii)

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

d²y/dx² =-(12)(3)sin3x

d²y/dx²=-(9)(4sin3x)

d²y/dx²=-9y (Since y=4sin3x)

d²y/dx2+9y=0

So, d²y/dx²+9y=0

Question 3: Show that y=ae^2x+be^−x is a solution of the differential equation.

d²y/dx²-dy/dx-2y=0

Solution:

y=ae^2x+be^−x (i)

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

dy/dx=2ae^2x-be^-x (ii)

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

d²y/dx²=4ae^2x+be^-x (iii)

we have,

d²y/dx²-dy/dx-2y (iv)

Putting the values ofandin equation (iv)

=4ae^2x+be^-x-(2ae^2x-be^-x)-2(ae^2x+be^−x)

=4ae^2x-4ae^2x +be^−x-be^−x)

=0

Question 4: Show that the function, y=Acosx-Bsinx is a solution of the differential equation.

d²y/dx²+y=0

Solution:

y=Acosx-Bsinx (i)

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

dy/dx=-Asinx-Bcosx (ii)

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

d²y/dx²=-Acosx+Bsinx

d²y/dx²=-(Acosx-Bsinx)

d²y/dx²+(Acosx-Bsinx)=0

d²y/dx²+y=0 (since y=Acosx-Bsinx)

Question 5: Show that the function, y=Acos2x-Bsin2x is a solution of the differential equation.

d²y/dx² + 4y = 0

Solution:

y=Acos2x-Bsin2x (i)

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

dy/dx=-2Asin2x-2Bcos2x (ii)

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

d²y/dx²=-4Acos2x+4Bsin2x

d²y/dx²+4(Acos2x-Bsin2x)=0

d²y/dx²+4y=0 (since y=Acos2x-Bsin2x)

Question 6: Show that, y=Ae^Bx is the solution of the differential equation.

d²y/dx²=(1/y)(dy/dx)²

Solution:

y=Ae^Bx (i)

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

dy/dx=ABe^Bx (ii)

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

d²y/dx²=AB²e^bx

d²y/dx²=(ABe^bx)² /(Ae^Bx)

d2y/dx2=(1/y)(dy/dx)²

Question 7: Verify that y= (x/a)+b is the solution of the differential equation.

d²y/dx²+(2/x)(dy/dx)²=0

Solution:

y= (x/a)+b (i)

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

dy/dx=-(a/x²) (ii)

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

d²y/dx²=+(2a/x³)

d²y/dx²=-(-2/x)(a/x²)

d²y/dx²+(2/x)(dy/dx)=0

Question 8: Verify that y²=4ax is the solution of the differential equation.

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

Solution:

y²=4ax (i)

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

2y(dy/dx)=4a

dy/dx=(2a/y)

we have,

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

=x(2a/y)+y(y/2a)

=(4xa+y²)/2y

=(2y²/2y)

=y

Question 9: Show that Ax²+By²=1 is the solution of the differential equation.

Solution:

Ax²+By²=1 (i)

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

2Ax+2By(dy/dx)=0

2Ax=-2By(dy/dx)

y(dy/dx)=-(Ax/B) (ii)

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

(dy/dx)²+y(d²y/dx²)=-(A/B)

(dy/dx)²+y(d²y/dx²)=-(y/x)(dy/dx)

Question 10: Show that y=ax³+bx²+cis the solution of the differential equation.

(d³y/dx³)=6a

Solution:

We have,

y=ax³+bx²+c (i)

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

(dy/dx)=3ax²+2bx (ii)

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

(d²y/dx²)=6ax (iii)

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

(d³y/dx³)=6a

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-1 x} 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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.