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

  • Last Updated : 18 Mar, 2021

Solve the following differential equations:

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 = e∫Pdx

= e∫2dx

= e2x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = ∫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 = e∫Pdx



= e∫2dx

= e2x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = (5/4)∫e-3x.e2xdx + c

y(e2x) = (5/4)∫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 = e∫Pdx

= e∫2dx

= e2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = ∫6ex.e2xdx + c



y(e2x) = 6∫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 = e∫Pdx

= e∫dx

= ex

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(ex) = ∫e-2x.exdx + c

y(ex) = ∫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 = e∫Pdx



= e-∫(dx/x)

= e-log(x)

= elog(1/x)

= (1/x)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(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 = e∫Pdx

= e∫2dx

= e2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = ∫4x.e2xdx + c

y(e2x) = 4x∫e2xdx – 4∫{(dx/dx)∫e2xdx}dx + c

y(e2x) = 2xe2x – 2∫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 = e∫Pdx

= e∫(dx/x)

= elog(x)

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.exdx + c

xy = x∫exdx – {(dx/dx)∫exdx}dx + c

xy = xex – ∫exdx + c

xy = xex – ex + c

y=\frac{x-1}{x}e^x+\frac{c}{x}

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 = e∫Pdx

=e^{∫\frac{4x}{x^2+1}dx}

=e^{2log|x^2+1|}

= (x2+1)2

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

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

y(x2 + 1)2 = -∫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 = e∫Pdx

= e∫(dx/x)

= elog(x)

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.logxdx + c

xy = logx∫xdx – {(d/dx)logx∫xdx}dx + c

xy = (x2/2)logx – ∫(1/x)(x2/2) + c

xy = (x2/2)logx – (1/2)∫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 = e∫Pdx

= e-∫(dx/x)

= e-log(x)

= elog(1/x)

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(\frac{1}{x})=∫(\frac{x-1}{x})e^x.(\frac{dx}{x})+c

(y/x) = ∫[(1/x) – (1/x2)]ex + c

Since, ∫[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 = e∫Pdx

= e∫dx/x

= elogx

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx = ∫x3.xdx + c

yx = ∫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 = e∫Pdx

= e∫dx

= ex

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(ex) = ∫sinx.exdx + c

Let, I = ∫sinx.exdx

I = ex∫sinxdx – ∫{(d/dx)ex∫sinxdx}dx

I = -excos + ∫excosxdx

I = -excosx + ex∫cosxdx – ∫{(d/dx)ex∫cosxdx}dx

I = -excosx + exsinx – ∫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 = e∫Pdx

= e∫dx

= ex

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(ex) = ∫cosx.exdx + c

Let, I = ∫cosx.exdx

I = ex∫cosxdx – ∫{(d/dx)ex∫cosxdx}dx

I = exsinx – ∫exsinxdx

I = exsinx – ex∫sinxdx + ∫{(d/dx)ex∫sinxdx}dx

I = exsinx + excosx – ∫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 = e∫Pdx

= e∫2dx

= e2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = ∫sinx.e2xdx + c

Let, I = ∫sinx.e2xdx

I = e2x∫sinxdx – {(d/dx)e2x∫sinxdx}dx

I = -e2xcosx + 2∫e2xcosdx

I = -e2xcosx + 2e2x∫cosxdx – 2{(d/dx)e2x∫cosxdx}dx

I = -e2xcosx + 2e2xsinx – 4∫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 = e∫Pdx

= e∫-tanxdx

= e-log|secx|

= 1/secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/secx) = -2∫sinx.(1/secx)dx + c

ycosx = -∫2sinx.cosxdx + c

ycosx = -∫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 = e∫Pdx

=e^{∫\frac{dx}{1+x^2}}



=e^{tan^{-1}x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{tan^{-1}x})=∫\frac{tan^{-1}x}{1+x^2}e^{tan^{-1}x}dx+c

Let, tan-1x = z

On differentiating both sides we get,

dx/(1 + x2) = dz

y(ez) = ∫zezdz + c

y(ez) = z∫ezdz – {(dz/dz)∫ezdz}dz

y(ez) = zez – ∫ ezdz + c

y(ez) = ez(z – 1) + c

y = (z – 1) + ce-z

y=(tan^{-1}x-1)+ce^{tan^{-1}x}

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 = e∫Pdx

= e∫tanxdx

= elog|secx|

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.secx = ∫cosx.secxdx + c

y.secx = ∫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 = e∫Pdx

= e∫cotxdx

= elog|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = ∫(x2cotx + 2x)sinxdx + c

y(sinx) = ∫(x2cosx + 2xsinx)dx + c

y(sinx) = x2∫cosxdx – {(d/dx)x2∫cosxdx}dx + ∫2xsinxdx + c

y(sinx) = x2sinx – ∫2xsinxdx + ∫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 = e∫Pdx

= e∫tanxdx

= elog|secx|



= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫secx.(x2.cos2x)dx + c

y(secx) = ∫x2.cosxdx + c

y(secx) = x2∫cosxdx – ∫{(d/dx)x2∫cosxdx}dx + c

y(secx) = x2sinx – 2∫xsinxdx + c

y(secx) = x2sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

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

y(secx) = x2sinx + 2xcosx – 2sinx + c

This is the required solution.

Question 20. (1 + x2)(dy/dx) + y = e^{tan^{-1}x}

Solution:

We have,

(1 + x2)(dy/dx) + y = e^{tan^{-1}x}

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

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(x2 + 1), Q = e^{tan^{-1}x}/(x2 + 1)

So,

I.F = e∫Pdx

=e^{∫\frac{1}{x^2+1}dx}

e^{tan^{-1}x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{tan^{-1}x}) – ∫[e^{tan^{-1}x}/(x2 + 1)]e^{tan^{-1}x}dx + c

Let, tan-1x = z

On differentiating both sides we get

dx/(1 + x2) = dz

yez = ∫e2zdz + c



yez = (e2z/2) + c

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

y = (1/2)e^{tan^{-1}x} + c.e^{tan^{-1}x}

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 = e∫Pdx

= e-2∫(dx/x)

= e-2log(x)

= e2log|1/x|

= 1/x2

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x2) = ∫(1/x2).(2x3 + x)dx + c

(y/x2) = ∫[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 – e^{tan^{-1}y})(dy/dx) = 0

Solution:

We have,

(1 + y2) + (x – e^{tan^{-1}y})(dy/dx) = 0

\frac{dy}{dx}=-\frac{1+y^2}{x-e^{tan^{-1}}x}

\frac{dx}{dy}=-\frac{x-e^{tan^{-1}}x}{1+y^2}

(dx/dy) + x/(1 + y2) = e^{tan^{-1}y}/(1 + y2)  ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/(1 + y2), Q = e^{tan^{-1}y}/(1 + y2)

So,

I.F = e∫Pdy

e^{∫\frac{dy}{1+y^2}}

e^{tan^{-1}y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c



x(e^{tan^{-1}y})=∫(e^{tan^{-1}y})(\frac{e^{tan^{-1}y}}{1+y^2})dy+c

Let, tan-1y = z

On differentiating both sides we have,

dy/(1 + y2) = dz

xez = ∫e2zdz + c

xez = (e2z/2) + c

x = ez/2 + ce-z

x=\frac{e^{tan^{-1}y}}{2}+c.e^{tan^{-1}y}

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 = e∫Pdy

= e∫dy/y2

= e-(1/y)

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

e-(1/y)x = ∫(1/y3).(e-1/y)dy + c

Let,-(1/y) = z

Differentiating both sides we have,

(dy/y2) = dz 

xez = -∫zezdz + c

xez = -z∫ezdz + ∫{(dz/dz)∫ezdz}dz + c

xez = -zez + ∫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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