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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.10 | Set 2

  • Last Updated : 18 Mar, 2021

Solve the following differential equations:

Question 24. (2x – 10y3)(dy/dx) + y = 0

Solution:

We have,

(2x – 10y3)(dy/dx) + y = 0

(2x – 10y3)(dy/dx) = -y

(dx/dy) = -(2x – 10y3)/y

(dx/dy) + 2x/y = 10y2  ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 2/y, Q = 10y2 

So,

I.F = e∫Pdy

= e∫(2/y)dy

= e2log|y|

= y2

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(y2) = ∫(10y2)(y2)dy + c

xy2 = 10(y5/5) + c

x = 2y3 + cy-2

This is the required solution.

Question 25. (x + tany)dy = sin2ydx

Solution:

We have,

(x + tany)dy = sin2ydx

(dx/dy) = (x + tany)/sin2y

(dx/dy) – cosec2y.x = tany/sin2y ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -cosec2y, Q = tany/sin2y

So, I.F = e∫Pdy

= e∫-cosec2ydy

e^{-\frac{1}{2}log|tany|}

e^{log|\sqrt{coty}|}

= √coty

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(√coty) = ∫(tany/sin2y).(√coty)dy + c

x\sqrt{coty}=∫\frac{\sqrt{tany}}{\frac{2tany}{1+tan^2y}}dy+c

x\frac{1}{\sqrt{tany}}=∫\frac{1+tan^2y}{2\sqrt{tany}}dy+c

x\frac{1}{\sqrt{tany}}=∫\frac{sec^2x}{2\sqrt{tany}}dy+c

Let, tany = z

On differentiating both side we have,

sec2ydx = dz

(x/√tany) = (1/2)∫dz/√z + c

(x/√tany) = (1/2)(2√z) + c

x = (√tany)(√tany) + c(√tany)

x = tany + c(√tany)

This is the required solution.

Question 26. dx + xdy = e-ysec2ydy

Solution:

We have,

dx + xdy = e-ysec2ydy

(x – e-ysec2y)dy = -dx

(dx/dy) = (e-ysec2y-x) ………..(i)

The above equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e-ysec2y

So, I.F = e∫Pdy

= e∫dy

= ey

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(ey) = ∫e-ysce2yeydy + c

x(ey) = ∫sec2ydy + c

x(ey) = tany + c

x = (tany + c)e-y

This is the required solution.

Question 27. (dy/dx) = ytanx – 2sinx

Solution:

We have,

(dy/dx) = ytanx – 2sinx

(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

= cosx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(cosx) = -2∫sinx.(cosx)dx + c

ycosx = -∫2sinx.cosxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

ycosx = -2∫zdz + c

ycosx = -2(z2/2) + c

ycosx = -sin2xdx + c

y = secx(-sin2xdx + c)

This is the required solution.

Question 28. (dy/dx) + ycosx = sinx.cosx

Solution:

We have,

(dy/dx) + ycosx = sinx.cosx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = sinx.cosx

So,

I.F = e∫Pdx

= e∫cosxdx

= esinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(esinx) = ∫(esinx)(sinx.cosx)dx + c

Let, sinx = z

Differentating both sides we get,

cosxdx = 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 = (sinx – 1) + ce-sinx

This is the required solution.

Question 29. (1 + x2)(dy/dx) – 2xy = (x2 + 2)(x2 + 1)

Solution:

We have,

(1 + x2)(dy/dx) – 2xy = (x2 + 2)(x2 + 1)

(dy/dx) – 2xy/(1 + x2) = (x2 + 2)  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

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

So, 

I.F = e∫Pdx

e^{-∫\frac{2x}{1+x^2}dx}

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

= 1/(x2+1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

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

y/(x2 + 1) = ∫[1 + 1/(x2 + 1)]dx + c

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

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

This is the required solution.

Question 30. sinx(dy/dx) + ycosx = 2sin2xcosx

Solution:

We have,

sinx(dy/dx) + ycosx = 2sin2xcosx

(dy/dx) + ycotx = 2sinx.cosx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2sinx.cosx

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) = ∫(2sinx.cosx)sinxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

y.z = 2∫z2 + c

y.z = (2/3)z3 + c

y.sinx = (2/3)sin3x + c

This is the required solution.

Question 31. (x2 – 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

Solution:

We have,

(x2 – 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

(dy/dx) + 2(x + 2)y/(x2 – 1) = 2(x + 1)/(x2 – 1)

(dy/dx) + 2(x + 2)y/(x2 – 1) = 2/(x – 1)   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2(x + 2)/(x2 – 1), Q = 2/(x – 1)

So,

I.F = e∫Pdx

e^{∫\frac{2(x+2)}{x^2-1}dx}

=e^{∫\frac{2x}{x^2-1}dx+∫\frac{4}{x^2-1}dx}

=e^{log|x^2-1|+\frac{4}{2}log|\frac{x-1}{x+1}|}

=e^{log(|x^2-1||\frac{x-1}{x+1}|^2)}

=e^{log|\frac{(x-1)^3}{x+1}|}

= (x – 1)3/(x + 1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[(x – 1)3/(x + 1)] = ∫[(x – 1)3/(x + 1)[{2/(x – 1)]dx + c

\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x-1)^2}{(x+1)}+c

\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x+1)^2-4x}{(x+1)}+c

\frac{(x-1)^3}{(x+1)}y=2∫[(x+1)-4\frac{x}{(x+1)}]+c

\frac{(x-1)^3}{(x+1)}y=∫[2x+2-8\frac{x}{x+1}]dx

y.[(x – 1)3/(x + 1)] = (x2 – 6x + 8log|x + 1|) + c

y=\frac{x+1}{(x-1)^3}(x^2-6x+8log|x+1|+c)

This is the required solution.

Question 32. (dy/dx) + (2y/x) = cosx

Solution:

We have,

(dy/dx) + (2y/x) = cosx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = cosx

So,

I.F = e∫Pdx

= e∫(2/x)dx

= e2log|x|

= x2 

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x2) = ∫(x2).(cosx)dx + c

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

x2y = x2sinx – 2∫xsinxdx + c

x2y = x2sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

x2y = x2sinx + 2xcosx – 2∫cosxdx + c

x2y = x2sinx + 2xcosx – 2sinx + c

This is the required solution.

Question 33. (dy/dx) – y = xex

Solution:

We have,

(dy/dx) – y = xex  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = xex

So,

I.F = e∫Pdx

= e∫-dx

= e-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-x) = ∫(e-x)(xex)dx + c

ye-x = ∫xdx + c

ye-x = (x2/2) + c

y = [(x2/2) + c].ex 

This is the required solution.

Question 34.  (dy/dx) + 2y = xe4x

Solution:

We have,

 (dy/dx) + 2y = xe4x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = xe4x

So,

I.F = e∫Pdx

= e2∫dx

= e2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e2x) = ∫(e2x).(xe4x)dx + c

y(e2x) = ∫xe6xdx + c

y(e2x) = x∫e6xdx – ∫{(dx/dx)∫e6xdx}dx + c

e2xy = (xe6x)/6 – ∫(e6x/6)dx + c

e2xy = (xe6x)/6 – e6x/36 + c

y = (xe4x)/6 – e4x/36 + ce-2x

This is the required solution.

Question 35. (x + 2y2)(dy/dx) = y, given that when x = 2, y = 1

Solution:

We have,

(x + 2y2)(dy/dx) = y

(dx/dy) = (x + 2y2)/y

(dx/dy) – (x/y) = 2y       ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y, Q = 2y

So,

I.F = e∫Pdy

= e∫-dy/y

= e-log|y|

= 1/y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(1/y) = ∫(1/y)(2y)dy + c

(x/y) = 2∫dy + c

(x/y) = 2y + c

x = 2y2 + cy

Given that when x = 2, y = 1

2 = 2 + c

c = 0

x = 2y2

This is the required solution.

Question 36(i). Find one-parameter families of solution curves of the following differential equation (dy/dx) + 3y = emx, m is a given real number

Solution:

We have,

(dy/dx) + 3y = emx      ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 3, Q = emx 

So,

I.F = e∫Pdx

= e∫3dx

= e3x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e3x) = ∫(e3x).(emx)dx + c

y(e3x) = ∫e(3+m)xdx + c

y(e3x) = e(m+3)x/(m + 3) + c

y = emx/(m + 3) + c

This is the required solution.

Question 36(ii). Find one-parameter families of solution curves of the following differential equation (dy/dx) – y = cos2x

Solution:

We have,

(dy/dx) – y = cos2x     ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = cos2x

So,

I.F = e∫Pdx

= e-∫dx

= e-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-x) = ∫(e-x).(cos2x)dx + c

y(e-x) = ∫e-xcos2xdx + c

Let, 

A = ∫e-xcos2xdx

= e-x∫cos2xdx – {(d/dx)e-x∫cos2xdx}dx

= (e-x/2)sin2x + ∫(e-x/2)sin2xdx

=e^{-x}sin2x+\frac{1}{2}e^{-x}∫sin2x-\frac{1}{2}∫[\frac{d}{dx}e^{-x}∫sin2xdx]

=\frac{1}{2}e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}∫e^{-x}cos2xdx

A=e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}A

(5/4)A = (e-x/2)(2sin2x – cos2x)

A=\frac{2}{5}e^{-x}(2sin2x-cos2x)

ye^{-x}=\frac{2}{5}e^{-x}(2sin2x-cos2x)+c

y=\frac{2}{5}(2sin2x-cos2x)+ce^x

This is the required solution.

Question 36(iii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) – y = (x + 1)e-x

Solution:

We have,

x(dy/dx) – y = (x + 1)e-x

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

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = [(x + 1)/x]e-x

So,

I.F = e∫Pdx

= e-∫dx/x

= e-log|x|

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫[(x+1)/x]e-x(1/x)dx + c

y/x = ∫[1/x+1/x2]e-xdx + c

Let, (1/x)e-x = z

On differentiating both sides we have

-[1/x + 1/x2]e-xdx = dz

y/x = -∫dz + c

y/x = -z + c

y/x = -(e-x/x) + c

y = -e-x + cx

This is the required solution.

Question 36(iv). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + y = x4 

Solution:

We have,

 x(dy/dx) + y = x

(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

= elog|x|

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(x)(x3)dx + c

xy = ∫x4 + c

xy = (x5/5) + c

y = (x4/5) + c/x

This is the required solution.

Question 36(v). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = logx

Solution:

We have,

(xlogx)(dy/dx) + y = logx

(dy/dx) + y/xlogx = 1/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 1/x

So,

I.F = e∫Pdx

= e∫dx/xlogx

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e∫dz/z

= elog|z|

= z

=l ogx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(1/x)(logx)dx + c

y(logx) = ∫zdz + c (Let, logx=z and differentiating both sides)

y(logx) = (z2/2) + c

y(logx) = (logx)2/2 + c

y = logx/2 + c/logx

This is the required solution.

Question 36(vi). Find one-parameter families of solution curves of the following differential equation (dy/dx) – 2xy/(1 + x2) = x2 + 2

Solution:

We have,

(dy/dx) – 2xy/(1 + x2) = x2 + 2   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

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

So,

I.F = e∫Pdx

= e-∫2xdx/(1+x2)

= e-log|1+x2|

= 1/(1+x2)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

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

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

\frac{y}{x^2+1}=∫\frac{x^2+1+1}{x^2+1}dx+c

y/(1 + x2) = ∫dx + ∫dx/(x2 + 1) + c

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

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

This is the required solution.

Question 36(vii). Find one-parameter families of solution curves of the following differential equation (dy/dx) + ycosx = esinxcosx

Solution:

We have,

(dy/dx) + ycosx = esinxcosx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = esinxcosx

So,

I.F = e∫Pdx

= e∫cosxdx

= esinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[(esinx) = ∫(esinx)(esinxcosx)dx + c

Let, sinx = z

ON differentiating both sides we have,

cosxdx = dz

yez = ∫e2zdz + c

yez = (e2z/2) + c

y = ez/2 + ce – z

y = (esinx/2) + ce-sinx

This is the required solution.

Question 36(viii). Find one-parameter families of solution curves of the following differential equation (x + y)(dy/dx) = 1

Solution:

We have,

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

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

(dx/dy) = (x + y)

(dx/dy) – x = y    ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -1, Q = y

So,

I.F = e∫Pdy

= e-∫dy

= e-y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e-y) = ∫(e-y)(y)dy + c

xe-y = y∫e-ydy – ∫{(dy/dy)∫e-ydy}dy + c

xe-y = -ye-y + ∫e-y + c

xe-y = -ye-y – e-y + c

xe-y + ye-y + e-y = c

e-y(x + y + 1) = c

(x + y + 1) = cey

This is the required solution.

Question 36(ix). Find one-parameter families of solution curves of the following differential equation cos2x(dy/dx) = (tanx – y)

Solution:

We have,

cos2x(dy/dx) = (tanx – y)

(dy/dx) = (tanx – y)/cos2x

(dy/dx) = tanx.sec2x – ysec2x

(dy/dx) + ysec2x = tanx.sec2x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = sec2x, Q = tanx.sec2x

So,

I.F = e∫Pdx

=e^{∫sec^2xdx}

= etanx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(etanx) = ∫(etanx)(tanx.sec2x)dx + c

Let, tanx = z

On differentiating both sides we have,

sec2xdx = dz

y(ez) = ∫zezdz + c

y(ez) = z∫ezdz – ∫{(dz/dz)∫ezdz}dz

y(ez) = zez – ∫ezdz + c

y(ez) = zez – ez + c

y = (z – 1) + c.e-z

y = (tanx – 1) + c.e-tanx

This is the required solution.

Question 36(x). Find one-parameter families of solution curves of the following differential equation e-ysec2ydy = dx + xdy

Solution:

We have,

dx + xdy = e-ysec2ydy

(x – e-ysec2y)dy = -dx

(dx/dy) = (e-ysec2y – x)

(dx/dy) + x = e-ysec2y        ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e-ysec2y

So, I.F = e∫Pdy

= e∫dy

= ey

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(ey) = ∫e-ysce2yeydy + c

x(ey) = ∫sec2ydy + c

x(ey) =tany + c

x = (tan y + c)e-y

This is the required solution.

Question 36(xi). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = 2logx

Solution:

We have,

(xlogx)(dy/dx) + y = 2logx

(dy/dx) + y/xlogx = 2/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 2/x

So, 

I.F = e∫Pdx

= e∫dx/xlogx

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e∫dz/z

= elog|z|

= z

= logx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(2/x)(logx)dx + c

y(logx) = 2∫zdz + c (Let, logx = z and differentiating both sides)

y(logx) = 2(z2/2) + c

y(logx) = (logx)2 + c

y = logx + c/logx

This is the required solution.

Question 36(xii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + 2y = x2logx

Solution:

We have,

x(dy/dx) + 2y = x2logx

(dy/dx) + 2y/x = xlogx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = xlogx

So,

I.F = e∫Pdx

= e2∫dx/x

= e2logx

= x2

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x2) = ∫(x2)(xlogx)dx + c

x2y = ∫x3logxdx + c

x2y = logx∫x3dx + ∫{(d/dx)logx∫x3dx}dx + c

x2y = (1/4)x4logx – (1/4)∫x3dx + c

x2y = (1/4)x4logx – (1/16)x4 + c 

y = (x2 /16)(4logx – 1) + c/x2

This is the required solution.

Question 37. Solve the following using the initial value problem:-

(i). y’ + y = ex, y(0) = (1/2)

Solution:

We have,

y’ + y = ex

dy/dx + y = ex    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = ex

So, I.F = e∫Pdx

= e∫dx

= ex

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(ex) = ∫ex.exdx + c

y(ex) = (1/2)e2x + c

At t = 0, y = (1/2)

(1/2)e0 = (1/2)e0 + c

c = 0

y(ex) = (1/2)e2x

y = (ex/2)

This is the required solution.

(ii). x(dy/dx) – y = logx, y(1) = 0

Solution:

We have,

x(dy/dx) – y = logx

(dy/dx) – y/x = logx/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = logx/x

So,

I.F = e∫Pdx

= e-∫dx/x

= e-log|x|

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(1/x)(logx/x)dx + c

(y/x) = ∫(logx/x2)dx + c

(y/x) = logx∫(dx/x2) – ∫{(d/dx)logx∫(dx/x2)}dx + c

(y/x) = -(logx/x) + ∫(dx/x2) + c

(y/x) = -(logx/x) – (1/x) + c

At x = 1, y = 0

0 = -0 – 1 + c

c = 1

(y/x) = -(logx/x) – (1/x) + 1

y = x – 1 – logx

This is the required solution.

(iii). (dy/dx) + 2y = e-2xsinx, y(0) = 0

Solution:

We have,

(dy/dx) + 2y = e-2xsinx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e-2xsinx

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) = ∫e-2xsinx.(e2x)dx + c

y(e2x) = ∫sinxdx + c

y(e2x) = -cosx + c

At x = 0, y = 0

0 = -1 + c

c = 1

y(e2x) = 1 – cosx

This is the required solution.

(iv). x(dy/dx) – y = (x + 1)e-x, y(1) = 0 

Solution:

We have,

x(dy/dx) – y = (x + 1)e-x

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

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x + 1)/x]e-x

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)]e-x + c

Since, -∫[f(x) + f'(x)]e-xdx = f(x)e-x + c

(y/x) = -e-x/x + c

At x = 1, y = 0

0 = -e-1 + c

c = e-1

(y/x) = -e-x/x + e-1

y = xe-1 – e-x

This is the required solution.

(v). (1 + y2)dx + (x – e^{-tan^{-1}y})dx = 0, y(0) = 0

Solution:

We have,

(1 + y2)(dx/dy) + x = e^{-tan^{-1}y}

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

The given equation is a linear differential equation of the form

(dy/dx) + Px = Q

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

So,

I.F = e∫Pdy

=e^{∫\frac{1}{y^2+1}dy}

= etan-1y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^{-tan^{-1}y}) = ∫[e^{-tan^{-1}y}/(y2 + 1)]e^{-tan^{-1}y}dx + c

x(e^{-tan^{-1}y}) = ∫dy/(1 + y2) + c

x(etan-1y) = tan-1y + c

At x = 0, y = 0

0*e0 = 0 + c

c = 0

x(e^{-tan^{-1}y}) = tan-1y

This is the required solution.

(vi). (dy/dx) + ytanx = x2tanx + 2x, y(0) = 1

Solution:

We have,

(dy/dx) + ytanx = x2tanx + 2x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x2tanx+2x

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) = ∫(x2tanx + 2x)secxdx + c

y(secx) = ∫(x2tanxsecx + 2xsecx)dx + c

y(secx) = ∫x2tanxsecxdx + 2∫xsecxdx + c

y(secx) = ∫x2tanxsecxdx + 2secx∫xdx – 2∫{(d/dx)secx∫xdx}dx + c

y(secx) = ∫x2tanxsecxdx + x2.secx – ∫x2tanxsecxdx + c

y(secx) = x2,(secx)+c

At, x = 0, y = 1

1 = 0 + c

c = 1

y = x2 + cosx

This is the required solution.

(vii). x(dy/dx) + y = xcosx + sinx, y(π/2) = 1

Solution:

We have,

x(dy/dx) + y = xcosx + sinx

(dy/dx) + (y/x) = cosx + sinx/x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = cosx + sinx/x

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) = ∫(cosx + sinx/x)(x)xdx + c

y(x) = ∫xcosxdx + ∫sinxdx + c

xy = x∫cosxdx – ∫{(dx/dx)∫cosxdx}dx – cosx + c

xy = xsinx – ∫sinxdx – cosx + c

xy = xsinx + cosx – cosx + c

xy = xsinx + c

At x = π/2, y = 1

π/2 = π/2sin(π/2) + c

c = 0

y = sinx

This is the required solution.

(viii). (dy/dx) + ycotx = 4xcosecx, y(π/2) = 0

Solution:

We have,

(dy/dx) + ycotx = 4xcosecx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 4xcosecx

So,

I.F = e∫Pdx

= e∫cotx.dx

= elog|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = 4∫(xcosecx)(sinx)xdx + c

y(sinx) = 4∫xdx + c

y(sinx) = 4(x2/2) + c

y(sinx) = 2x2 + c

At x = π/2, y = 0,

0 = 2(π/2)2 + c

c = -π2/2

y(sinx) = 2x2 – π2/2

This is the required solution.

(ix). (dy/dx) + 2ytanx = sinx, y = 0 when x = π/3

Solution:

We have,

(dy/dx) + 2ytanx = sinx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2tanx, Q = sinx

So, I.F = e∫Pdx

= e∫2tanxdx

= e2log|secx|

= sec2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.sec2x = ∫sinx.sec2xdx + c

y.sec2x = ∫tanx.secxdx + c

y.sec2x = secx+c

At x = π/3, y = 0,

0 = sec2(π/3) + c

c = -2

y.sec2x = secx – 2

y = cosx – 2cos2x

This is the required solution.

(x). (dy/dx) – 3ycotx = sin2x, y = 2 when x = π/2

Solution:

We have,

(dy/dx) – 3ycotx = sin2x        ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -3cotx, Q = sin2x

So, I.F = e∫Pdx

= e∫-3cotxdx

= e-3log|sinx|

= cosec3x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(cosec3x) = 2∫(cosec3x).(sin2x)dx + c

y.(cosec3x) = 2∫cotx.cosecxdx + c

y.(cosec3x) = -2cosesx +c

y = -2sin2x + c.sin3x

At x = π/2, y = 2.

2 = -2sin2(π/2) + c.sin3(π/2)

c = 4

y = c.sin3x – 2sin2x

This is the required solution.

(xi). (dy/dx) + ycotx = 2cosx, y(π/2) = 0 

Solution:

We have,

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

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx 

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) = ∫(sinx).(2cosx)dx + c

y.(sinx) = 2∫sinx.cosxdx + c

y.(sinx) = ∫sin2xdx + c

y.(sinx) = -(cos2x/2) + c

At x = π/2, y = 0

0 = -cos(π)/2 + c

c = -(1/2)

y.(sinx) = -(cos2x/2) – (1/2)

2y(sinx) = -(1 + cos2x)

2y(sinx) = -2cos2x

y = -cotx.cosx

This is the required solution.

(xii). dy = cosx(2 – ycosecx)dx,

Solution:

We have,

dy = cosx(2 – ycosecx)dx

(dy/dx) = -ycotx + 2cosx

(dy/dx) + ycotx = 2cosx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx

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) = 2∫cosx.(sinx)dx + c

ysinx = ∫2cosx.sinxdx + c

ysinx = ∫sin2x + c

ysinx = -(cos2x/2) + c

This is the required solution.

Question 38. x(dy/dx) + 2y = x2

Solution:

We have,

x(dy/dx) + 2y = x2

(dy/dx) + 2y/x = x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = x

So, I.F = e∫Pdx

= e2∫dx/x

= e2logx

= x2

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx2 = ∫x2.xdx + c

yx2 = ∫x3dx + c

x2y = (x4/4) + c

y = (x1/4) + c.x-2

This is the required solution.

Question 39. (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

= e-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e-x) = ∫cosx.e-xdx + c

Let, I = ∫cosx.e-xdx

I = e-x∫cosxdx – ∫{(d/dx)e-x∫cosxdx}dx

I = e-xsinx + ∫e-xsinxdx

I = e-xsinx + e-x∫sinxdx-∫{(d/dx)e-x∫sinxdx}dx

I = e-xsinx – e-xcosx-∫e-xcosxdx

2I = e-x(sinx – cosx)

I = (e-x/2)(sinx – cosx)

y(e-x) = (e-x/2)(sinx – cosx) + c

y = (1/2)(sinx-cosx) + cex

This is the required solution.

Question 40. (y + 3x2)(dx/dy) = x

Solution:

We have,

(y + 3x2)(dx/dy) = x

(dy/dx) = (y + 3x2)/x

(dy/dx) – y/x = 3x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = 3x

So, I.F = e∫Pdx

= e-∫dx/x

= e-logx

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = 3∫x.(1/x)dx + c

y(1/x) = 3∫dx + c

y/x = 3x + c

This is the required solution.

Question 41. Find a particular solution of the differential equation (dx/dy) + xcoty = y2coty + 2y, given that x = 0, when y = π/2

Solution:

We have,

(dx/dy) + xcoty = y2coty + 2y     ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = coty, Q = y2coty + 2y

So,

I.F = e∫Pdy

= e∫cotydy

= elog|siny|

= siny

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(siny) = ∫(y2coty + 2y)sinydy + c

x(siny) = ∫(y2cosy + 2xsiny)dy + c

x(siny) = y2∫cosydx – {(d/dy)y2∫cosydy}dy + ∫2ysinydy + c

x(siny) = y2siny – ∫2ysinydy + ∫2ysinydy + c

x(siny) = y2siny + c

At x = 0, y = π/2

0 = (π/2)2sin(π/2) + c

c = -π2/4

x(siny) = y2siny – π2/4

This is the required solution.


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