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

### Solve the following differential equations:

### Question 24. (2x – 10y^{3})(dy/dx) + y = 0

**Solution:**

We have,

(2x – 10y

^{3})(dy/dx) + y = 0(2x – 10y

^{3})(dy/dx) = -y(dx/dy) = -(2x – 10y

^{3})/y(dx/dy) + 2x/y = 10y

^{2}………..(i)

The given equation is a linear differential equation of the form(dx/dy) + Px = Q

Where, P = 2/y, Q = 10y

^{2}So,

I.F = e

^{∫Pdy}= e

^{∫(2/y)dy}= e

^{2log|y|}= y

^{2}The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(y

^{2}) = ∫(10y^{2})(y^{2})dy + cxy

^{2}= 10(y^{5}/5) + cx = 2y

^{3 }+ 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}=

=

= √coty

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

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

Let, tany = z

On differentiating both side we have,

sec

^{2}ydx = 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^{-y}sec^{2}ydy

**Solution:**

We have,

dx + xdy = e

^{-y}sec^{2}ydy(x – e

^{-y}sec^{2}y)dy = -dx(dx/dy) = (e

^{-y}sec^{2}y-x)………..(i)The above equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e

^{-y}sec^{2}ySo, 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}sce^{2}ye^{y}dy + cx(e

^{y}) = ∫sec^{2}ydy + cx(e

^{y}) = tany + cx = (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(z

^{2}/2) + cycosx = -sin

^{2}xdx + cy = secx(-sin

^{2}xdx + 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}= e

^{sinx}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{sinx}) = ∫(e^{sinx})(sinx.cosx)dx + cLet, sinx = z

Differentating both sides we get,

cosxdx = dz

y(e

^{z}) = ∫ze^{z}dz + cy(e

^{z}) = z∫e^{z}dz – {(dz/dz)∫e^{z}dz}dzy(e

^{z}) = ze^{z }– ∫e^{z}dz + cy(e

^{z}) = e^{z}(z – 1) + cy = (z – 1) + ce

^{-z}y = (sinx – 1) + ce

^{-sinx}This is the required solution.

### Question 29. (1 + x^{2})(dy/dx) – 2xy = (x^{2 }+ 2)(x^{2 }+ 1)

**Solution:**

We have,

(1 + x

^{2})(dy/dx) – 2xy = (x^{2 }+ 2)(x^{2 }+ 1)(dy/dx) – 2xy/(1 + x

^{2}) = (x^{2 }+ 2)………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = -2x/(1 + x

^{2}), Q = (x^{2 }+ 1)So,

I.F = e

^{∫Pdx}=

= 1/(x

^{2}+1)The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[1/(x

^{2 }+ 1)] = ∫[(x^{2 }+ 2)/(x^{2 }+ 1)]dx + cy/(x

^{2 }+ 1) = ∫[1 + 1/(x^{2 }+ 1)]dx + cy/(x

^{2 }+ 1) = x + tan^{-1}x + cy = (x

^{2 }+ 1)(x + tan^{-1}x + c)This is the required solution.

### Question 30. sinx(dy/dx) + ycosx = 2sin^{2}xcosx

**Solution:**

We have,

sinx(dy/dx) + ycosx = 2sin

^{2}xcosx(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}= e

^{log|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∫z

^{2 }+ cy.z = (2/3)z

^{3 }+ cy.sinx = (2/3)sin

^{3}x + cThis is the required solution.

### Question 31. (x^{2 }– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

**Solution:**

We have,

(x

^{2 }– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)(dy/dx) + 2(x + 2)y/(x

^{2 }– 1) = 2(x + 1)/(x^{2}– 1)(dy/dx) + 2(x + 2)y/(x

^{2 }– 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)/(x

^{2 }– 1), Q = 2/(x – 1)So,

I.F = e

^{∫Pdx}=

= (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 + cy.[(x – 1)

^{3}/(x + 1)] = (x^{2 }– 6x + 8log|x + 1|) + cThis 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}= e

^{2log|x|}= x

^{2}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x

^{2}) = ∫(x^{2}).(cosx)dx + cx

^{2}(y) = x^{2}∫cosxdx – ∫{(d/dx)x^{2}∫cosxdx}dx + cx

^{2}y = x^{2}sinx – 2∫xsinxdx + cx

^{2}y = x^{2}sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + cx

^{2}y = x^{2}sinx + 2xcosx – 2∫cosxdx + cx

^{2}y = x^{2}sinx + 2xcosx – 2sinx + cThis is the required solution.

### Question 33. (dy/dx) – y = xe^{x}

**Solution:**

We have,

(dy/dx) – y = xe

^{x}………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = -1, Q = xe

^{x}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})(xe^{x})dx + cye

^{-x }= ∫xdx + cye

^{-x }= (x^{2}/2) + cy = [(x

^{2}/2) + c].e^{x}This is the required solution.

### Question 34. (dy/dx) + 2y = xe^{4x}

**Solution:**

We have,

(dy/dx) + 2y = xe

^{4x}………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = 2, Q = xe

^{4x}So,

I.F = e

^{∫Pdx}= e

^{2∫dx}= e

^{2x}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{2x}) = ∫(e^{2x}).(xe^{4x})dx + cy(e

^{2x}) = ∫xe^{6x}dx + cy(e

^{2x}) = x∫e^{6x}dx – ∫{(dx/dx)∫e^{6x}dx}dx + ce

^{2x}y = (xe^{6x})/6 – ∫(e^{6x}/6)dx + ce

^{2x}y = (xe^{6x})/6 – e^{6x}/36 + cy = (xe

^{4x})/6 – e^{4x}/36 + ce^{-2x}This is the required solution.

### Question 35. (x + 2y^{2})(dy/dx) = y, given that when x = 2, y = 1

**Solution:**

We have,

(x + 2y

^{2})(dy/dx) = y(dx/dy) = (x + 2y

^{2})/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 = 2y

^{2 }+ cyGiven that when x = 2, y = 1

2 = 2 + c

c = 0

x = 2y

^{2}This is the required solution.

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

**Solution:**

We have,

(dy/dx) + 3y = e

^{mx}………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = 3, Q = e

^{mx}So,

I.F = e

^{∫Pdx}= e

^{∫3dx}= e

^{3x}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{3x}) = ∫(e^{3x}).(e^{mx})dx + cy(e

^{3x}) = ∫e^{(3+m)x}dx + cy(e

^{3x}) = e^{(m+3)x}/(m + 3) + cy = e

^{mx}/(m + 3) + cThis 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 + cy(e

^{-x}) = ∫e^{-x}cos2xdx + cLet,

A = ∫e

^{-x}cos2xdx= e

^{-x}∫cos2xdx – {(d/dx)e^{-x}∫cos2xdx}dx= (e

^{-x}/2)sin2x + ∫(e^{-x}/2)sin2xdx=

(5/4)A = (e

^{-x}/2)(2sin2x – cos2x)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 + cy/x = ∫[1/x+1/x

^{2}]e^{-x}dx + cLet, (1/x)e

^{-x }= zOn differentiating both sides we have

-[1/x + 1/x2]e

^{-x}dx = dzy/x = -∫dz + c

y/x = -z + c

y/x = -(e

^{-x}/x) + cy = -e

^{-x }+ cxThis is the required solution.

### Question 36(iv). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + y = x^{4}

**Solution:**

We have,

x(dy/dx) + y = x

^{4 }(dy/dx) + y/x = x

^{3}………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = 1/x, Q = x

^{3}So,

I.F = e

^{∫Pdx}= e

^{∫dx/x}= e

^{log|x|}= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(x)(x

^{3})dx + cxy = ∫x

^{4 }+ cxy = (x

^{5}/5) + cy = (x

^{4}/5) + c/xThis 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}= e

^{log|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) = (z

^{2}/2) + cy(logx) = (logx)

^{2}/2 + cy = 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 + x^{2}) = x^{2 }+ 2

**Solution:**

We have,

(dy/dx) – 2xy/(1 + x

^{2}) = x^{2 }+ 2………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = -2x/(1 + x

^{2}), Q = x^{2 }+ 2So,

I.F = e

^{∫Pdx}= e

^{-∫2xdx/(1+x2)}= e

^{-log|1+x2|}= 1/(1+x

^{2})The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[1/(1 + x

^{2})] = ∫[1/(1 + x^{2})](x^{2 }+ 2)dx + cy/(1 + x

^{2}) = ∫[(x^{2 }+ 2)/(x^{2 }+ 1)]dx + cy/(1 + x

^{2}) = ∫dx + ∫dx/(x^{2 }+ 1) + cy/(x

^{2 }+ 1) = x + tan^{-1}x + cy = (x + tan

^{-1}x + c)(x^{2 }+ 1)This is the required solution.

### Question 36(vii). Find one-parameter families of solution curves of the following differential equation (dy/dx) + ycosx = e^{sinx}cosx

**Solution:**

We have,

(dy/dx) + ycosx = e

^{sinx}cosx………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = cosx, Q = e

^{sinx}cosxSo,

I.F = e

^{∫Pdx}= e∫cosxdx

= e

^{sinx}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[(e

^{sinx}) = ∫(e^{sinx})(e^{sinx}cosx)dx + cLet, sinx = z

ON differentiating both sides we have,

cosxdx = dz

ye

^{z }= ∫e^{2z}dz + cye

^{z }= (e^{2z}/2) + cy = ez/2 + ce – z

y = (e

^{sinx}/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 + cxe

^{-y }= y∫e^{-y}dy – ∫{(dy/dy)∫e^{-y}dy}dy + cxe

^{-y }= -ye^{-y }+ ∫e^{-y }+ cxe

^{-y }= -ye^{-y }– e^{-y }+ cxe

^{-y }+ ye^{-y }+ e^{-y }= ce

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

^{y}This is the required solution.

### Question 36(ix). Find one-parameter families of solution curves of the following differential equation cos^{2}x(dy/dx) = (tanx – y)

**Solution:**

We have,

cos

^{2}x(dy/dx) = (tanx – y)(dy/dx) = (tanx – y)/cos

^{2}x(dy/dx) = tanx.sec

^{2}x – ysec^{2}x(dy/dx) + ysec

^{2}x = tanx.sec^{2}x………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = sec

^{2}x, Q = tanx.sec^{2}xSo,

I.F = e∫Pdx

= e

^{tanx}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{tanx}) = ∫(e^{tanx})(tanx.sec^{2}x)dx + cLet, tanx = z

On differentiating both sides we have,

sec

^{2}xdx = dzy(e

^{z}) = ∫ze^{z}dz + cy(e

^{z}) = z∫e^{z}dz – ∫{(dz/dz)∫e^{z}dz}dzy(e

^{z}) = ze^{z }– ∫e^{z}dz + cy(e

^{z}) = ze^{z }– e^{z }+ cy = (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^{-y}sec^{2}ydy = dx + xdy

**Solution:**

We have,

dx + xdy = e

^{-y}sec^{2}ydy(x – e

^{-y}sec^{2}y)dy = -dx(dx/dy) = (e

^{-y}sec^{2}y – x)(dx/dy) + x = e

^{-y}sec^{2}y………..(i)

The given equation is a linear differential equation of the form(dx/dy) + Px = Q

Where, P = 1, Q = e

^{-y}sec^{2}ySo, 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}sce^{2}ye^{y}dy + cx(e

^{y}) = ∫sec^{2}ydy + cx(e

^{y}) =tany + cx = (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}= e

^{log|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(z

^{2}/2) + cy(logx) = (logx)

^{2 }+ cy = 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 = x^{2}logx

**Solution:**

We have,

x(dy/dx) + 2y = x

^{2}logx(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

= e

^{2∫dx/x}= e

^{2logx}= x

^{2}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x

^{2}) = ∫(x^{2})(xlogx)dx + cx

^{2}y = ∫x^{3}logxdx + cx

^{2}y = logx∫x^{3}dx + ∫{(d/dx)logx∫x^{3}dx}dx + cx

^{2}y = (1/4)x^{4}logx – (1/4)∫x^{3}dx + cx

^{2}y = (1/4)x^{4}logx – (1/16)x^{4 }+ cy = (x

^{2 }/16)(4logx – 1) + c/x^{2}This is the required solution.

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

### (i). y’ + y = e^{x}, y(0) = (1/2)

**Solution:**

We have,

y’ + y = e

^{x}dy/dx + y = e

^{x}………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = 1, Q = e

^{x}So, I.F = e

^{∫Pdx}= e

^{∫dx}= e

^{x}The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{x}) = ∫e^{x}.e^{x}dx + cy(e

^{x}) = (1/2)e^{2x }+ cAt t = 0, y = (1/2)

(1/2)e

^{0 }= (1/2)e^{0 }+ cc = 0

y(e

^{x}) = (1/2)e^{2x}y = (e

^{x}/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/x

^{2})dx + c(y/x) = logx∫(dx/x

^{2}) – ∫{(d/dx)logx∫(dx/x^{2})}dx + c(y/x) = -(logx/x) + ∫(dx/x

^{2}) + 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^{-2x}sinx, y(0) = 0

**Solution:**

We have,

(dy/dx) + 2y = e

^{-2x}sinx………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = 2, Q = e

^{-2x}sinxSo,

I.F = e

^{∫Pdx}= e

^{∫2dx}= e

^{2x}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e

^{2x}) = ∫e^{-2x}sinx.(e^{2x})dx + cy(e

^{2x}) = ∫sinxdx + cy(e

^{2x}) = -cosx + cAt x = 0, y = 0

0 = -1 + c

c = 1

y(e

^{2x}) = 1 – cosxThis 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)}= e

^{log(1/x)}= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

(y/x) = ∫[(1/x) + (1/x

^{2})]e^{-x }+ cSince, -∫[f(x) + f'(x)]e

^{-x}dx = f(x)e^{-x }+ c(y/x) = -e

^{-x}/x + cAt x = 1, y = 0

0 = -e

^{-1 }+ cc = e

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

^{-x}/x + e^{-1}y = xe

^{-1 }– e^{-x}This is the required solution.

### (v). (1 + y^{2})dx + (x – )dx = 0, y(0) = 0

**Solution:**

We have,

(1 + y

^{2})(dx/dy) + x =(dy/dx) + [1/(y

^{2 }+ 1)]x = /(y^{2 }+ 1)………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Px = Q

Where, P = 1/(y

^{2 }+ 1), Q = /(y^{2 }+ 1)So,

I.F = e

^{∫Pdy}= e

^{tan-1y}The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x() = ∫[/(y

^{2 }+ 1)]dx + cx() = ∫dy/(1 + y

^{2}) + cx(e

^{tan-1y}) = tan^{-1}y + cAt x = 0, y = 0

0*e

^{0 }= 0 + cc = 0

x() = tan

^{-1}yThis is the required solution.

### (vi). (dy/dx) + ytanx = x^{2}tanx + 2x, y(0) = 1

**Solution:**

We have,

(dy/dx) + ytanx = x

^{2}tanx + 2x………..(i)

The given equation is a linear differential equation of the form(dy/dx) + Py = Q

Where, P = tanx, Q = x

^{2}tanx+2xSo,

I.F = e

^{∫Pdx}= e

^{∫tanxdx}= e

^{log|secx|}= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫(x

^{2}tanx + 2x)secxdx + cy(secx) = ∫(x

^{2}tanxsecx + 2xsecx)dx + cy(secx) = ∫x

^{2}tanxsecxdx + 2∫xsecxdx + cy(secx) = ∫x

^{2}tanxsecxdx + 2secx∫xdx – 2∫{(d/dx)secx∫xdx}dx + cy(secx) = ∫x

^{2}tanxsecxdx + x^{2}.secx – ∫x^{2}tanxsecxdx + cy(secx) = x

^{2},(secx)+cAt, x = 0, y = 1

1 = 0 + c

c = 1

y = x

^{2 }+ cosxThis 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}= e

^{log|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}= e

^{log|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(x

^{2}/2) + cy(sinx) = 2x

^{2 }+ cAt x = π/2, y = 0,

0 = 2(π/2)

^{2 }+ cc = -π

^{2}/2y(sinx) = 2x

^{2 }– π^{2}/2This 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}= e

^{2log|secx|}= sec

^{2}xThe solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.sec

^{2}x = ∫sinx.sec^{2}xdx + cy.sec

^{2}x = ∫tanx.secxdx + cy.sec

^{2}x = secx+cAt x = π/3, y = 0,

0 = sec

^{2}(π/3) + cc = -2

y.sec

^{2}x = secx – 2y = cosx – 2cos

^{2}xThis 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|}= cosec

^{3}xThe solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(cosec

^{3}x) = 2∫(cosec^{3}x).(sin2x)dx + cy.(cosec

^{3}x) = 2∫cotx.cosecxdx + cy.(cosec

^{3}x) = -2cosesx +cy = -2sin

^{2}x + c.sin^{3}xAt x = π/2, y = 2.

2 = -2sin

^{2}(π/2) + c.sin^{3}(π/2)c = 4

y = c.sin

^{3}x – 2sin^{2}xThis 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}= e

^{log|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) = -2cos

^{2}xy = -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}= e

^{log|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 = x^{2}

**Solution:**

We have,

x(dy/dx) + 2y = x

^{2}(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}= e

^{2∫dx/x}= e

^{2logx}= x

^{2}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx

^{2 }= ∫x2.xdx + cyx

^{2 }= ∫x^{3}dx + cx

^{2}y = (x^{4}/4) + cy = (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^{-x}dx + cLet, I = ∫cosx.e

^{-x}dxI = e

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

^{-x}sinx + ∫e^{-x}sinxdxI = e

^{-x}sinx + e^{-x}∫sinxdx-∫{(d/dx)e^{-x}∫sinxdx}dxI = e

^{-x}sinx – e^{-x}cosx-∫e^{-x}cosxdx2I = e

^{-x}(sinx – cosx)I = (e

^{-x}/2)(sinx – cosx)y(e

^{-x}) = (e^{-x}/2)(sinx – cosx) + cy = (1/2)(sinx-cosx) + ce

^{x}This is the required solution.

### Question 40. (y + 3x^{2})(dx/dy) = x

**Solution:**

We have,

(y + 3x

^{2})(dx/dy) = x(dy/dx) = (y + 3x

^{2})/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 = y^{2}coty + 2y, given that x = 0, when y = π/2

**Solution:**

We have,

(dx/dy) + xcoty = y

^{2}coty + 2y………..(i)

The given equation is a linear differential equation of the form(dx/dy) + Px = Q

Where, P = coty, Q = y

^{2}coty + 2ySo,

I.F = e

^{∫Pdy}= e

^{∫cotydy}= e

^{log|siny|}= siny

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(siny) = ∫(y

^{2}coty + 2y)sinydy + cx(siny) = ∫(y

^{2}cosy + 2xsiny)dy + cx(siny) = y

^{2}∫cosydx – {(d/dy)y^{2}∫cosydy}dy + ∫2ysinydy + cx(siny) = y

^{2}siny – ∫2ysinydy + ∫2ysinydy + cx(siny) = y

^{2}siny + cAt x = 0, y = π/2

0 = (π/2)

^{2}sin(π/2) + cc = -π

^{2}/4x(siny) = y

^{2}siny – π^{2}/4This is the required solution.