Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e)

I.

Find the I.F. of the following differential equations by transforming them into linear form.

Question 1.

x \(\frac{d y}{d x}\) – y = 2x^{2} sec^{2} 2x.

Solution:

The given equation can be expressed as

\(\frac{d y}{d x}-\frac{y}{x}\) = 2x sec^{2} 2x

This is of the form \(\frac{d y}{d x}\) + Py = Q where the

Integrating factor I.F = e^{∫ P dx}, P = – \(\frac{1}{x}\)

= e^{– ∫ \(\frac{1}{x}\) dx}

= e^{– log x}

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

Question 2.

y \(\frac{d y}{d x}\) – x = 2y^{3}

Solution:

The given equation can be written as

\(\frac{d x}{d y}-\frac{x}{y}\) = 2y^{2}

and the integrating factor I.F = e^{∫ P dy}

= e^{– ∫ \(\frac{1}{y}\) dy}

= e^{– log y}

= e^{log y-1} = \(\frac{1}{y}\).

II. Solve the following differential equations.

Question 1.

\(\frac{d y}{d x}\) + y tan x = cos^{3} x

Solution:

Given \(\frac{d y}{d x}\) + y tan x = cos^{3} x

which is of the form \(\frac{d y}{d x}\) + Py = Q where

P = tan x and Q = cos^{3} x

∴ Integrating Factor I.F. = e^{∫ P dx}

= e^{∫ tan x dx}

= e^{log sec x} = sec x

General solution is y. sec x = ∫ Q (I.F.) dx

= ∫ cos^{3} x sec x dx

= ∫ cos^{2} x dx

= \(\int \frac{1+\cos 2 x}{2} d x=\frac{1}{2} x+\frac{1}{4} \sin 2 x\)

= \(\frac{1}{2}\) (x + sin x cos x) + c

\(\frac{y}{\cos x}\) = \(\frac{1}{2}\) (x + sin x cos x) + c

⇒ 2y = cos x (x + sin x cos x) + c cos x

= x cos x + sin x cos^{2} x + c cos x is the solution.

Question 2.

\(\frac{d y}{d x}\) + y sec x = tan x

Solution:

This is of the form \(\frac{d y}{d x}\) + Py = Q

where P = sec x and Q = tan x

∴ Integrating Factor I.F. = e^{∫ sec x dx}

= e^{log (sec x + tan x)}

= sec x + tan x

∴ General solution is

y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c

∴ y (sec x + tan x) = ∫ tan x (sec x + tan x) + c

= ∫ sec x tan x dx . ∫ tan^{2} x dx

= sec x + ∫ (sec^{2} x – 1) dx + c

= sec x + tan x – x + c

∴ y (sec x + tan x) sec x + tan x – x + c is the solution.

Question 3.

\(\frac{d y}{d x}\) – y tan x = e^{x} sec x.

Solution:

This is of the form \(\frac{d y}{d x}\) + Py = Q where

P = – tan x and Q = e^{x} sec x.

∴ Integrating Factor IF. = e^{∫ P dx} dx

= e^{∫ tan x} dx

= e^{log cos x} = cos x

∴ General solution is y . e^{∫ P dx} dx = ∫ Q . e^{∫ P dx} dx + c

∴ y . cos x = ∫ e^{x} sec x cos x dx + c

= ∫ e^{x} dx + c

= e^{x} dx + c

y = e^{x} sec x + c sec x. is the solution.

Question 4.

x \(\frac{d y}{d x}\) + 2y = log x.

Solution:

The equation can be written as

\(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2}{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{x}}{\mathrm{x}}\)

This is of the form \(\frac{d y}{d x}\) + Py = Q where

I.F. = e^{∫ P dx} where P = \(\frac{2}{x}\), and Q = \(\frac{\pi}{2}\)

= e^{∫ \(\frac{2}{x}\)}

= e^{log x2} = x^{2}.

y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c

Question 5.

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

Solution:

The equation can be written as

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

Question 6.

\(\frac{d y}{d x}+\frac{2 y}{x}\) = 2x^{2}

Solution:

The given equation can be written as

\(\frac{d y}{d x}+\frac{2 y}{x}\) = 2x^{2}

∴ I.F.= e^{∫ P dx}

= e^{∫ \(\frac{2}{x}\) dx}

= e^{2 log x}

= e^{log x2} = x^{2}

∴ Solution is y . x^{2} = ∫ 2x^{2} . x^{2} dx

= 2 ∫ x^{4} dx

= 2 \(\frac{x^5}{5}\) + c.

Question 7.

\(\frac{d y}{d x}+\frac{4 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)

Solution:

This is of the form \(\frac{d y}{d x}\) + Py = Q where

P = \(\frac{4 x}{1+x^2}\) and Q = \(\frac{1}{\left(1+x^2\right)^2}\)

∴ I.F = e^{∫ P dx}

= \(e^{\int \frac{4 x}{1+x^2} d x}\)

= e^{2 log (1 + x2)}

= (1 + x^{2})^{2} dx + c

= x + c is the solution.

Question 8.

x \(\frac{d y}{d x}\) + y = (1 + x) e^{x}

Solution:

The given equation can be written as

\(\frac{d y}{d x}+\frac{y}{x}=\frac{1+x}{x} e^x\)

∴ I.F = e^{∫ P dx}

= e^{∫ \(\frac{1}{x}\) dx}

= e^{log x} = x

∴ Solution is

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

= ∫ (1 + x) e^{x} dx + c

= ∫ e^{x} dx + ∫ x e^{x} dx + c

= e^{x} + x e^{x} – e^{x} + c

= x e^{x} + c is the solution.

Question 9.

\(\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}\)

Solution:

Question 10.

\(\frac{d y}{d x}\) – y = – 2 e^{-x}

Solution:

Here P = – 1 and Q = – 2 e^{-x}

∴ I.F. = e^{∫ P dx} dx

= e^{∫ – 1 dx} dx

∴ Solution is

y . e^{-x} = ∫ – 2 e^{-x} e^{-x} dx

= – 2 ∫ e^{-2x} dx

= \(\frac{(-2)}{(-2)}\) e^{-2x} + c

= e^{-2x} + c

∴ y = e^{-x} + C e^{x} is the solution.

Question 11.

(1 + x^{2}) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = Tan^{-1} x

Solution:

The equation can be written as

= ∫ t e^{t} dt where t = tan^{-1} x

= t e^{t} – e^{t}

= e^{tan-1 x (tan-1 – 1)} + c

y = (tan^{-1} x – 1) + c e^{– tan-1 x} is the solution.

Question 12.

\(\frac{d y}{d x}\) + y tan x = sin x

Solution:

We have P = tan x and Q = sin x

∴ I.F = e^{∫ P dx} dx

= e^{∫ tan x dx} dx

= e^{log sec x} dx = sec x

∴ Solution is

y sec x = ∫ sin x . sec x dx + c

= ∫ tan x dx

= log |sec x| + c is the solution.

III. Solve the following differential equations.

Question 1.

cos x + y sin x = sec^{2} x

Solution:

The given equation can be written as

\(\frac{d y}{d x}\) + sin x sec x =sec x

Here P = sin x sec x – tan x and Q = sec^{3} x

∴ I.F. = e^{∫ P dx} dx

= e^{∫ tan x dx} dx

= e^{log (sec x)} = sec x

∴ Solution is

y sec x = ∫ sec^{4} x dx + c

= ∫ (1 + tan^{2} x) sec^{2} x dx + c

= tan x + \(\frac{\tan ^3 x}{3}\) + c.

Question 2.

sec x dy = (y + sin x) dx

Solution:

The given equation can be written as

sec x \(\frac{d y}{d x}\) = y + sin x

⇒ \(\frac{d y}{d x}=\frac{y}{\sec x}+\frac{\sin x}{\sec x}\)

= y cos x + sin x cos x

\(\frac{d y}{d x}\) – y cos x = sin x cos x

I.F. = e^{∫ P dx}

= e^{– ∫ cos x dx}

= e^{– sin x}

∴ Solution is y e^{– sin x}

= ∫ sin x cos x e^{– sin x} dx

= ∫ t e^{– t} dt where t = sin x

= t (- e^{– t}) + ∫ e^{– t} dt

= – e^{– t} (t + 1) + c

= – e^{– sin x} (sin x + 1) + c

y = – (sin x + 1) + c e^{sin x} is the solution.

Question 3.

x log x \(\frac{d y}{d x}\) + y = 2 log x

Solution:

The equation can be written as

Question 4.

(x + y + 1) \(\frac{d y}{d x}\) = 1

Solution:

From the given equation

\(\frac{d y}{d x}=\frac{1}{x+y+1}\)

\(\frac{d y}{d x}\) = x + y + 1

∴ \(\frac{d y}{d x}\) – x = (y + 1)

This is of the form \(\frac{d x}{d y}\) + x = Q(y)

P = – 1 and Q = (y + 1)

∴ I.F. = e^{∫ P dy} = e^{– y}

∴ Solution is

x e^{– y} = ∫ (y + 1) e^{– y} dy

= ∫ e^{– y} y dy + ∫ e^{– y} dy

= – y e^{– y} + ∫ e^{– y} dy – e^{– y}

= – y e^{– y} – e^{– y} – e^{– y} + c

= – y e^{– y} – 2 e^{– y} + c

x = – y – 2 + ce^{y}

= – (y + 2) ce^{y} is the solution.

Question 5.

x (x – 1) \(\frac{d y}{d x}\) – y = x^{3} (x – 1)^{3}

Solution:

The equation can be written as

Question 6.

(x + 2y^{3}) \(\frac{d y}{d x}\) = y.

Solution:

The given differential equation is

(x + 2y^{3}) \(\frac{d y}{d x}\) = y

Question 7.

(1 – x^{2}) \(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-\mathbf{x}^2}\)

Solution:

Dividing by (1 – x^{2}) both sides

Question 8.

x (x – 1) \(\frac{d y}{d x}\) – (x – 2) y = x^{3} (2x – 1)

Solution:

Dividing the given equation by x(x – 1) we get

\(\frac{d y}{d x}+\frac{(-x+2)}{x(x-1)} y=\frac{x^2(2 x-1)}{x-1}\)

Now \(\frac{2-x}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\)

∴ 2 – x = A (x – 1) + Bx

Put x = 1 both sides, 1 = B

and A + B = – 1

⇒ A = – 2

∴ \(\frac{2-x}{x(x-1)}=\frac{-2}{x}+\frac{1}{x-1}\)

∴ I.F = \(e^{\int \frac{2-x}{x(x-1)} d x}=e^{\int\left(\frac{-2}{x}+\frac{1}{x-1}\right) d x}\)

= e^{log (x – 1) – 2 log x}

= e^{log (x – 1) – log x2}

= \(\frac{x-1}{x^2}\)

∴ General Solution is \(y\left(\frac{x-1}{x^2}\right)=\int \frac{x^2(2 x-1)}{x-1}\left(\frac{x-1}{x^2}\right) d x\)

= ∫ (2x – 1) dx + c

= x^{2} – x + c

∴ y (x – 1) = x^{2} (x^{2} – x + c) is the solution of the differential equation.

Question 9.

\(\frac{d y}{d x}\) (x^{2} y^{3} + xy) = 1

Solution:

Question 10.

\(\frac{d y}{d x}\) + x sin 2y = x^{3} cos^{2} y

Solution:

Dividing by cos^{2} y we get

sec^{2} y \(\frac{d y}{d x}\) + 2x tan y = x^{3} …………….(1)

[∴ \(\frac{\sin 2 y}{\cos ^2 y}=\frac{2 \sin y \cos y}{\cos ^2 y}\) = 2 tan y]

Let tan y = t then sec^{2} y \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)

∴ \(\frac{d t}{d x}\) + 2xt = x^{3}

Here, P = 2x and Q = x^{3}

∴ I.F = e^{∫ P dx} = e^{x2}

∴ t . e^{x2} = ∫ x^{3} e^{x2} dx

= ∫ x^{2} . x . e^{x2} dx

= \(\frac{1}{2}\) ∫ z . e^{z} dz

where z = x^{2}

= \(\frac{1}{2}\) e^{z} (z – 1) + c

= \(\frac{1}{2}\) e^{x2} (x^{2} – 1) + c

∴ tan y e^{x2} = \(\frac{1}{2}\) e^{x2} (x^{2} – 1) + c

∴ Solution of the given differential equation is tan y = \(\frac{1}{2}\) (x^{2} – 1) + c e-^{x2}

Question 11.

y^{2} + (x – \(\frac{1}{y}\)) \(\frac{d y}{d x}\) = 0.

Solution:

Where P = y^{-2} and Q = y^{-3}

∴ I.F. = e^{∫ P dx}

= e∫^{y-2 dy}

= \(e^{-\frac{1}{y}}\)

∴ Solution is

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

= ∫ e^{-y-1} y^{-3} dy + c

= ∫ e^{-y-1} y^{-2} y^{-1} dy + c

Let y^{-1} = t then – y^{-2} dy = dt

∴ x \(e^{-\frac{1}{y}}\) = – ∫ t e^{-t} dt + c

= – [- t e^{-t} + ∫ e^{-t} dt] + c

= t e^{-t} + e^{-t} + c

= e^{-t} (t + 1) + c

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

∴ x = (\(\frac{1}{y}\) + 1) + c \(e^{\frac{1}{y}}\)

⇒ xy = 1 + y + y . c\(e^{\frac{1}{y}}\)

∴ Solution of the given differential equation is xy = 1 + y + cy \(e^{\frac{1}{y}}\).