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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.3 | Set 1
  • Last Updated : 03 Mar, 2021
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Question 1: Show that y=bex+ce2x is the solution of the differential equation.

d2 y/dx2-3(dy/dx)+2y=0

Solution:

y=bex+ce2x (i)

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

dy/dx=bex +2ce2x



dy/dx=bex+2ce2x (ii)

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

d2y/dx2 =bex+4ce2x (iii)

we have,

d2y/dx2 -3(dy/dx)+2y=0 (iv)

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

=bex+4ce2x-3(be2x+2ce2x)+2(bex+ce2x)

=3bex-3bex+6ce2x-6ce2x



=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=ae2x+be−x is a solution of the differential equation.

d2y/dx2-dy/dx-2y=0

Solution:

y=ae2x+be−x (i)

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

dy/dx=2ae2x-be-x (ii)

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

d2y/dx2=4ae2x+be-x (iii)

we have,

d2y/dx2-dy/dx-2y (iv)

Putting the values of\frac{d^2y}{dx^2} and\frac{dy}{dx} in equation (iv)

=4ae2x+be-x-(2ae2x-be-x)-2(ae2x+be−x)

=4ae2x-4ae2x +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=AeBx is the solution of the differential equation.

d2y/dx2=(1/y)(dy/dx)2

Solution:

y=AeBx (i)

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

dy/dx=ABeBx (ii)

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

d2y/dx2=AB2ebx

d2y/dx2=(ABebx)2 /(AeBx)

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.

x[y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2]=y\frac{dy}{dx}

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)

x[(\frac{dy}{dx})^2+y\frac{d^2y}{dx^2}]=y(\frac{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

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