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# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.10 | Set 1

• Last Updated : 18 Mar, 2021

### Question 1. dy/dx + 2y = e3x

Solution:

We have,

dy/dx + 2y = e3x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e3x

So, I.F = eâˆ«Pdx

= eâˆ«2dx

= e2x

The solution of differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(e2x) = âˆ«e3x.e2xdx + c

y(e2x) = (1/5)e5x + c

y = (e3x/5) + ce-2x

Hence, this is the required solution.

### Question 2. 4(dy/dx) + 8y = 5e-3x

Solution:

We have,

4(dy/dx) + 8y = 5e-3x

(dy/dx) + 2y = (5/4)e-3x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = (5/4)e-3x

So, I.F = eâˆ«Pdx

= eâˆ«2dx

= e2x

The solution of differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(e2x) = (5/4)âˆ«e-3x.e2xdx + c

y(e2x) = (5/4)âˆ«e-xdx + c

y = -(5/4)e-3x + ce-2x

This is the required solution.

### Question 3. (dy/dx) + 2y = 6ex

Solution:

We have,

(dy/dx) + 2y = 6ex ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 6ex

So, I.F = eâˆ«Pdx

= eâˆ«2dx

= e2x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(e2x) = âˆ«6ex.e2xdx + c

y(e2x) = 6âˆ«e3xdx + c

y(e2x) = 2e3x + c

y = 2ex + ce-2x

This is the required solution.

Question 4. (dy/dx) + y = e-2x

Solution:

We have,

(dy/dx) + y = e-2x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = e-2x

So, I.F = eâˆ«Pdx

= eâˆ«dx

= ex

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(ex) = âˆ«e-2x.exdx + c

y(ex) = âˆ«e-xdx + c

y(ex) = -e-x + c

y = -e-2x + ce-x

This is the required solution.

### Question 5. x(dy/dx) = x + y

Solution:

We have,

x(dy/dx) = x + y

(dy/dx) = 1 + (y/x)

(dy/dx) – (y/x) = 1  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (-1/x), Q = 1

So, I.F = eâˆ«Pdx

= e-âˆ«(dx/x)

= e-log(x)

= elog(1/x)

= (1/x)

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(1/x) = âˆ«(1/x)dx + c

(y/x) = log|x| + c

y = xlog|x| + cx

This is the required solution.

### Question 6. (dy/dx) + 2y = 4x

Solution:

We have,

(dy/dx) + 2y = 4x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 4x

So,

I.F = eâˆ«Pdx

= eâˆ«2dx

= e2x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(e2x) = âˆ«4x.e2xdx + c

y(e2x) = 4xâˆ«e2xdx – 4âˆ«{(dx/dx)âˆ«e2xdx}dx + c

y(e2x) = 2xe2x – 2âˆ«e2xdx + c

y(e2x) = 2xe2x – e2x + c

y = (2x – 1) + ce-2x

This is the required solution.

### Question 7. x(dy/dx) + y = xex

Solution:

We have,

x(dy/dx) + y = xex

(dy/dx) + (y/x) = ex  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = ex

So,

I.F = eâˆ«Pdx

= eâˆ«(dx/x)

= elog(x)

= x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(x) = âˆ«x.exdx + c

xy = xâˆ«exdx – {(dx/dx)âˆ«exdx}dx + c

xy = xex – âˆ«exdx + c

xy = xex – ex + c

This is the required solution.

### Question 8. (dy/dx) + [4x/(x2 + 1)]y = -1/(x2 + 1)2

Solution:

We have,

(dy/dx) + [4x/(x2 + 1)]y = -1/(x2 + 1)2    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = [4x/(x2 + 1)], Q = -1/(x2 + 1)2

So,

I.F = eâˆ«Pdx

= (x2+1)2

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(x2 + 1)2 = âˆ«-[1/(x2 + 1)2](x2 + 1)2dx + c

y(x2 + 1)2 = -âˆ«dx + c

y(x2 + 1)2 = -x + c

y = -x/(x2 + 1)2 + c/(x2 + 1)2

This is the required solution.

### Question 9. x(dy/dx) + y = xlogx

Solution:

We have,

x(dy/dx) + y = xlogx

(dy/dx) + (y/x) = logx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = logx

So,

I.F = eâˆ«Pdx

= eâˆ«(dx/x)

= elog(x)

= x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(x) = âˆ«x.logxdx + c

xy = logxâˆ«xdx – {(d/dx)logxâˆ«xdx}dx + c

xy = (x2/2)logx – âˆ«(1/x)(x2/2) + c

xy = (x2/2)logx – (1/2)âˆ«xdx + c

xy = (x2/2)logx – (x2/4) + c

y = (x/2)logx – (x/4) + (c/x)

This is the required solution.

### Question 10. x(dy/dx) – y = (x – 1)ex

Solution:

We have,

x(dy/dx) – y = (x – 1)ex

(dy/dx) – (y/x) = [(x – 1)/x]ex  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x – 1)/x]ex

So,

I.F = eâˆ«Pdx

= e-âˆ«(dx/x)

= e-log(x)

= elog(1/x)

= 1/x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

(y/x) = âˆ«[(1/x) – (1/x2)]ex + c

Since, âˆ«[f(x) + f'(x)]exdx = f(x)ex + c

(y/x) = (ex/x) + c

y = ex + xc

This is the required solution.

### Question 11. (dy/dx) + y/x = x3

Solution:

We have,

(dy/dx) + y/x = x3     ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x3

So, I.F = eâˆ«Pdx

= eâˆ«dx/x

= elogx

= x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

yx = âˆ«x3.xdx + c

yx = âˆ«x4dx + c

yx = (x5/5) + c

y = (x4/5) + c/x

This is the required solution.

### Question 12. (dy/dx) + y = sinx

Solution:

We have,

(dy/dx) + y = sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = sinx

So, I.F = eâˆ«Pdx

= eâˆ«dx

= ex

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(ex) = âˆ«sinx.exdx + c

Let, I = âˆ«sinx.exdx

I = exâˆ«sinxdx – âˆ«{(d/dx)exâˆ«sinxdx}dx

I = -excos + âˆ«excosxdx

I = -excosx + exâˆ«cosxdx – âˆ«{(d/dx)exâˆ«cosxdx}dx

I = -excosx + exsinx – âˆ«exsinxdx

2I = ex(sinx – cosx)

I = (ex/2)(sinx – cosx)

y(ex) = (ex/2)(sinx – cosx) + c

y = (1/2)(sinx – cosx) + ce-x

This is the required solution.

### Question 13. (dy/dx) + y = cosx

Solution:

We have,

(dy/dx) + y = cosx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = cosx

So, I.F = eâˆ«Pdx

= eâˆ«dx

= ex

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(ex) = âˆ«cosx.exdx + c

Let, I = âˆ«cosx.exdx

I = exâˆ«cosxdx – âˆ«{(d/dx)exâˆ«cosxdx}dx

I = exsinx – âˆ«exsinxdx

I = exsinx – exâˆ«sinxdx + âˆ«{(d/dx)exâˆ«sinxdx}dx

I = exsinx + excosx – âˆ«excosxdx

2I = ex(cosx + sinx)

I = (ex/2)(cosx + sinx)

y(ex) = (ex/2)(cosx + sinx) + c

y = (1/2)(cosx + sinx) + ce-x

This is the required solution.

### Question 14. (dy/dx) + 2y = sinx

Solution:

We have,

(dy/dx) + 2y = sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = sinx

So, I.F = eâˆ«Pdx

= eâˆ«2dx

= e2x

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(e2x) = âˆ«sinx.e2xdx + c

Let, I = âˆ«sinx.e2xdx

I = e2xâˆ«sinxdx – {(d/dx)e2xâˆ«sinxdx}dx

I = -e2xcosx + 2âˆ«e2xcosdx

I = -e2xcosx + 2e2xâˆ«cosxdx – 2{(d/dx)e2xâˆ«cosxdx}dx

I = -e2xcosx + 2e2xsinx – 4âˆ«e2xsinxdx

5I = e2x(2sinx – cosx)

I = (e2x/5)(2sinx – cosx)

y(e2x) = (e2x/5)(2sinx – cosx) + c

y = (1/5)(2sinx – cosx) + ce-2x

This is the required solution.

### Question 15. (dy/dx) – ytanx = -2sinx

Solution:

We have,

(dy/dx) – ytanx = -2sinx ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -tanx, Q = sinx

So,

I.F = eâˆ«Pdx

= eâˆ«-tanxdx

= e-log|secx|

= 1/secx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(1/secx) = -2âˆ«sinx.(1/secx)dx + c

ycosx = -âˆ«2sinx.cosxdx + c

ycosx = -âˆ«sin2xdx + c

ycosx = (cos2x/2) + c

y = (cos2x/cosx) + (c/cosx)

This is the required solution.

### Question 16. (1 + x2)(dy/dx) + y = tan-1x

Solution:

We have,

(1 + x2)(dy/dx) + y = tan-1x

(dy/dx) + [y/(1 + x2)] = tan-1x/(1 + x2)  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(1 + x2), Q = tan-1x/(1 + x2)

So,

I.F = eâˆ«Pdx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

Let, tan-1x = z

On differentiating both sides we get,

dx/(1 + x2) = dz

y(ez) = âˆ«zezdz + c

y(ez) = zâˆ«ezdz – {(dz/dz)âˆ«ezdz}dz

y(ez) = zez – âˆ« ezdz + c

y(ez) = ez(z – 1) + c

y = (z – 1) + ce-z

This is the required solution.

### Question 17. (dy/dx) + ytanx = cosx

Solution:

We have,

(dy/dx) + ytanx = cosx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = cosx

So, I.F = eâˆ«Pdx

= eâˆ«tanxdx

= elog|secx|

= secx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y.secx = âˆ«cosx.secxdx + c

y.secx = âˆ«dx + c

y.secx = x + c

y = xcosx + c.cosx

This is the required solution.

### Question 18. (dy/dx) + ycotx = x2cotx + 2x

Solution:

We have,

(dy/dx) + ycotx = x2cotx + 2x ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = x2cotx + 2x

So,

I.F = eâˆ«Pdx

= eâˆ«cotxdx

= elog|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(sinx) = âˆ«(x2cotx + 2x)sinxdx + c

y(sinx) = âˆ«(x2cosx + 2xsinx)dx + c

y(sinx) = x2âˆ«cosxdx – {(d/dx)x2âˆ«cosxdx}dx + âˆ«2xsinxdx + c

y(sinx) = x2sinx – âˆ«2xsinxdx + âˆ«2xsinxdx + c

y(sinx) = x2sinxdx + c

This is the required solution.

### Question 19. (dy/dx) + ytanx = x2cos2x

Solution:

We have,

(dy/dx) + ytanx = x2cos2x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x2cos2x

So,

I.F = eâˆ«Pdx

= eâˆ«tanxdx

= elog|secx|

= secx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(secx) = âˆ«secx.(x2.cos2x)dx + c

y(secx) = âˆ«x2.cosxdx + c

y(secx) = x2âˆ«cosxdx – âˆ«{(d/dx)x2âˆ«cosxdx}dx + c

y(secx) = x2sinx – 2âˆ«xsinxdx + c

y(secx) = x2sinx – 2xâˆ«sinxdx + 2âˆ«{(dx/dx)âˆ«sinxdx}dx + c

y(secx) = x2sinx + 2xcosx – 2âˆ«cosxdx + c

y(secx) = x2sinx + 2xcosx – 2sinx + c

This is the required solution.

Question 20. (1 + x2)(dy/dx) + y =

Solution:

We have,

(1 + x2)(dy/dx) + y =

(dy/dx) + [1/(x2 + 1)]y = /(x2 + 1)  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(x2 + 1), Q = /(x2 + 1)

So,

I.F = eâˆ«Pdx

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y() – âˆ«[/(x2 + 1)]dx + c

Let, tan-1x = z

On differentiating both sides we get

dx/(1 + x2) = dz

yez = âˆ«e2zdz + c

yez = (e2z/2) + c

y = (ez/z) + c.e-z

y = (1/2) + c.

This is the required solution.

### Question 21. xdy = (2y + 2x4 + x2)dx

Solution:

We have,

xdy = (2y + 2x4 + x2)dx

(dy/dx) = 2(y/x) + 2x3 + x

(dy/dx) – (y/x) = 2x3 + x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(2/x), Q = 2x3 + x

So,

I.F = eâˆ«Pdx

= e-2âˆ«(dx/x)

= e-2log(x)

= e2log|1/x|

= 1/x2

The solution of a differential equation is,

y(I.F) = âˆ«Q(I.F)dx + c

y(1/x2) = âˆ«(1/x2).(2x3 + x)dx + c

(y/x2) = âˆ«[2x + (1/x)]dx + c

(y/x2) = x2 + log|x| + c

y = x4 + x2log|x| + cx2

This is the required solution.

### Question 22. (1 + y2) + (x – )(dy/dx) = 0

Solution:

We have,

(1 + y2) + (x – )(dy/dx) = 0

(dx/dy) + x/(1 + y2) = /(1 + y2)  ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/(1 + y2), Q = /(1 + y2)

So,

I.F = eâˆ«Pdy

The solution of a differential equation is,

x(I.F) = âˆ«Q(I.F)dy + c

Let, tan-1y = z

On differentiating both sides we have,

dy/(1 + y2) = dz

xez = âˆ«e2zdz + c

xez = (e2z/2) + c

x = ez/2 + ce-z

This is the required solution.

### Question 23. y2(dx/dy) + x – 1/y = 0

Solution:

We have,

y2(dx/dy) + x – 1/y = 0

(dx/dy) + (x/y2) = 1/y3    ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y2, Q = 1/y3

So,

I.F = eâˆ«Pdy

= eâˆ«dy/y2

= e-(1/y)

The solution of a differential equation is,

x(I.F) = âˆ«Q(I.F)dy + c

e-(1/y)x = âˆ«(1/y3).(e-1/y)dy + c

Let,-(1/y) = z

Differentiating both sides we have,

(dy/y2) = dz

xez = -âˆ«zezdz + c

xez = -zâˆ«ezdz + âˆ«{(dz/dz)âˆ«ezdz}dz + c

xez = -zez + âˆ«ezdz + c

xez = -zez + ez + c

x = (1 – z) + ce – z

x = [1 + (1/y)] + ce1/y

x = (y + 1)/y + ce1/y

This is the required solution.

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