# 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=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/âˆš(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/âˆš(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,

(âˆšx2+a2)d2y/dx2+(1/(2âˆšx2+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=Â±âˆš(a2-x2), Differential equation: x+y(dy/dx)=0

Solution:

we have,

y=Â±âˆš(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)*(xÂ±a)2, Differential equation, y=(dy/dx)2

Solution:

We have,

y=(1/4)*(xÂ±a)2

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

dy/dx=(1/4)*2(xÂ±a)

Squaring both side, we have

(dy/dx)2=(1/4)*(xÂ±a)2

(dy/dx)2=y

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