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

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 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 = elog 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 = elog 1/y = \(\frac{1}{y}\)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}\) + y tan x = cos³ x
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 1
Solution is 2y = x cos x + sin x. cos² x + c. cos x

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

Question 2.
\(\frac{dy}{dx}\) + y sec x = tan x
Solution:
I.F. = e∫sec x dx = elog(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 = elog cos x = cos x
y. cos x = ∫ex.sec x. cos x dx = ∫ ex dx
= ex + c

Question 4.
x\(\frac{dy}{dx}\) + 2y = log x.
Solution:
I.F. = e∫\(\frac{2}{x}\)dx = e2log x = elog x² = x²
Solution is
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 2

Question 5.
(1 + x²)\(\frac{dy}{dx}\) + y = etan-1 x
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 3
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 4

Question 6.
\(\frac{dy}{dx}+\frac{2y}{x}\) = 2x².
Solution:
I.F. = e∫\(\frac{2}{x}\)dx = e2log x = elog x² = x²
y. x² = ∫2x4 dx = \(\frac{2x^5}{5}\) + c

Question 7.
\(\frac{dy}{dx}+\frac{4x}{1+x^2}y=\frac{1}{(1+x^2)^2}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 5

Question 8.
x\(\frac{dy}{dx}\) + y = (1 + x)ex
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 6

Question 9.
\(\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{1+x^2}{1+x^3}\).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 7

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. ex

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

Question 11.
(1 + x²)\(\frac{dy}{dx}\) + y = tan-1 x.
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 8
Put t = tan-1 x so that dt = \(\frac{dx}{1+x^2}\)
R.H.S = ∫t. etdt = t. et – ∫etdt
= t. et – et = et(t – 1)
Solution is y. etan-1 x = etan-1 x (tan-11 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 = elog 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 = elog sec x = sec x
y. sec x = ∫sec4 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 = + ∫et t dt
= t. et – et + 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. esin x.

Question 3.
x log x.\(\frac{dy}{dx}\) + y = 2 log x.
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 9

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
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 10

Question 6.
(x + 2y³)\(\frac{dy}{dx}\) = y
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 11
Solution is x = y(y² + c)

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

Question 7.
(1 – x²)\(\frac{dy}{dx}\) + 2xy = x\(\sqrt{1-x^2}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 12

Question 8.
x(x – 1)\(\frac{dy}{dx}\) – (x – 2)y = x³(2x – 1)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 13
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 14
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
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 15
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 16

Question 10.
\(\frac{dy}{dx}\) + x sin 2y = x³ cos² y
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 17
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 18

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

Question 11.
y² + (1 – \(\frac{1}{y}\)).\(\frac{dy}{dx}\) = 0
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 19
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e) 20
Solution xy = 1 + y + cy.e1/y

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