Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e)

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

Question 1.

x\(\frac{dy}{dx}\) – y = 2x² sec² 2x

Solution:

x\(\frac{dy}{dx}\) – y = 2x² sec² 2x

\(\frac{dy}{dx}\) – \(\frac{1}{x}\) . y = 2x. sec² 2x

This is linear in x

I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log x} = e^{log 1/x} = \(\frac{1}{x}\)

Question 2.

y\(\frac{dy}{dx}\) – x = 2y³

Solution:

y\(\frac{dy}{dx}\) – x = 2y³

\(\frac{dy}{dx}\) – \(\frac{1}{y}\).x = 2y²

I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log y} = e^{log 1/y} = \(\frac{1}{y}\)

II. Solve the following differential equations.

Question 1.

\(\frac{dy}{dx}\) + y tan x = cos³ x

Solution:

Solution is 2y = x cos x + sin x. cos² x + c. cos x

Question 2.

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

Solution:

I.F. = e^{∫sec x dx} = e^{log(secx + tan x)} = sec x + tan x

y. (sec x + tan x)

= ∫tan x (sec x + tan x)dx

= ∫(sec x. tan x + tan² x)dx

= ∫(sec x tan x + sec² x – 1)dx

Solution is

y(sec x + tan x) = sec x + tan x – x + c

Question 3.

\(\frac{dy}{dx}\) – y tan x = e sec x. dx

Solution:

I.F. = e^{-∫tan x dx} = e^{log cos x} = cos x

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

= e^{x} + c

Question 4.

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

Solution:

I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²

Solution is

Question 5.

(1 + x²)\(\frac{dy}{dx}\) + y = e^{tan-1 x}

Solution:

Question 6.

\(\frac{dy}{dx}+\frac{2y}{x}\) = 2x².

Solution:

I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²

y. x² = ∫2x^{4} dx = \(\frac{2x^5}{5}\) + c

Question 7.

\(\frac{dy}{dx}+\frac{4x}{1+x^2}y=\frac{1}{(1+x^2)^2}\)

Solution:

Question 8.

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

Solution:

Question 9.

\(\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{1+x^2}{1+x^3}\).

Solution:

Question 10.

\(\frac{dy}{dx}\) – y = -2e^{-x}.

Solution:

I.F = e^{∫-dx} = e^{-x}

y. e^{-x} = -2∫e^{-2x} dx = e^{-2x} + c

y = e^{-x} + c. e^{x}

Question 11.

(1 + x²)\(\frac{dy}{dx}\) + y = tan^{-1} x.

Solution:

Put t = tan^{-1} x so that dt = \(\frac{dx}{1+x^2}\)

R.H.S = ∫t. e^{t}dt = t. e^{t} – ∫e^{t}dt

= t. e^{t} – e^{t} = e^{t}(t – 1)

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

y = tan^{-1} x – 1 + c. e^{-tan-1 x}

Question 12.

\(\frac{dy}{dx}\) + y tan x = sin x.

Solution:

I.F. =e^{∫tan x dx} = e^{log sec x} = sec x

y. sec x = ∫ sin x. sec x dx = ∫tan x dx

= log sec x + c

III. Solve the following differential equations.

Question 1.

cos x. \(\frac{dy}{dx}\) + y sin x = sec² x

Solution:

\(\frac{dy}{dx}\) + tan x. y = sec³ x

I.F. = e^{∫tan x dx} = e^{log sec x} = sec x

y. sec x = ∫sec^{4} x dx = ∫(1 + tan² x) sec² x

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

Question 2.

sec x. dy = (y + sin x) dx.

Solution:

\(\frac{dy}{dx}=\frac{y+sin x}{sec x}\) = y cos x + sin x. cos x

\(\frac{dy}{dx}\) – y. cos x = sin x. cos x

I.F. = e^{-∫cos x dx} = e^{– sin x}

= y. e^{-sin x} = ∫ e^{-sin x}. sin x. cos x dx ……….. (1)

Consider ∫e^{-sin x}. sin x. cos x dx

t = – sin x ⇒ dt = -cos x dx

∫e^{-sin x}. sin x. cos x dx = + ∫e^{t} t dt

= t. e^{t} – e^{t} + c.

= e^{-sin x} (- sin x – 1) + c

y. e^{-sin x} = – e^{-sin x} (sin x + 1) + c

or y = – (sin x + 1) + c. e^{sin x}.

Question 3.

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

Solution:

Question 4.

(x + y + 1)\(\frac{dy}{dx}\) = 1.

Solution:

\(\frac{dy}{dx}\) = x + y + 1

\(\frac{dy}{dx}\) – x = y + 1

I.F. = e^{∫-dy} e^{-y}

x . e^{-y} = ∫e^{-y} (y + 1) dy

= -(y + 1). e^{-y} + ∫e^{-y}. dy

= -(y + 1) e^{-y} – e^{-y}

= -(y + 2) e^{-y}+ c

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

Question 5.

x(x – 1)\(\frac{dy}{dx}\) – y = x³(x – 1)³

Solution:

\(\frac{dy}{dx}-\frac{1}{x(x-1)}\)y = x²(x – 1)²

Question 6.

(x + 2y³)\(\frac{dy}{dx}\) = y

Solution:

Solution is x = y(y² + c)

Question 7.

(1 – x²)\(\frac{dy}{dx}\) + 2xy = x\(\sqrt{1-x^2}\)

Solution:

Question 8.

x(x – 1)\(\frac{dy}{dx}\) – (x – 2)y = x³(2x – 1)

Solution:

Solution is y(x – 1) = x²(x² – x + c)

Question 9.

\(\frac{dy}{dx}\)(x²y³ + xy) = 1

Solution:

\(\frac{dy}{dx}\) = xy + x²y³

This is Bernoulli’s equation

Question 10.

\(\frac{dy}{dx}\) + x sin 2y = x³ cos² y

Solution:

Question 11.

y² + (1 – \(\frac{1}{y}\)).\(\frac{dy}{dx}\) = 0

Solution:

Solution xy = 1 + y + cy.e^{1/y}