# RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations

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## RD Sharma Class 12 Ex 22.3 Solutions Chapter 22 Differential Equations

### 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/dx=bex+4ce2x (iii)

we have,

d2y/dx2 -3(dy/dx)+2y=0 (iv)

Putting the values ofdy/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=4sin3xis 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/dx=-(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−xis 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 and 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.

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=ex(Acosx+Bsinx) is the solution of the differential equation.

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

Solution:

we have,

y=ex(Acosx+Bsinx)               (i)

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

dy/dx=ex(Acosx+Bsinx)+ex(-Asinx+Bcosx)              (ii)

dy/dx=ex[(A+B)cosx-(A-B)sinx]                   (iii)

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

d2y/dx2 =ex(Acosx+Bsinx)+ex(-Asinx+Bcosx)+ex(-Asinx+Bcosx)+ex(-Acosx-Bsinx)

d2y/dx2=2ex[Bcosx-Asinx]          (iv)

d2y/dx2=2ex[(A+B)cosx-(A-B)sinx] -2ex(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=cetan-1 x is a solution of the differential equation.

(1+x2)(d2y/dx2)+(2x-1)(dy/dx)=0

Solution:

we have,

y=cetan-1 x                 (i)

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

dy/dx=cetan-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=em cos-1x is a solution of the differential equation.

(1-x2)(d2y/dx2)-x(dy/dx)-m2y=0

Solution:

we have,

y=em 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))             (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 isthe 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.

exd2y/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

exd2y/dx2=1

### (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)

=(-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

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