Solve the following differential equation: x d y + y d x = x 1 - x 2 y 2 d x

Given, x d y + y d x = x 1 - x 2 y 2 d x ⇒ x d y + y d x 1 - x 2 y 2 = x d x ....(1) Let u = sin - 1 x y Differentiating both the sides, d u = 1 1 - x y 2 d x y [Using chain rule of differentiation] ⇒ d u = 1 1 - x 2 y 2 x d y + y d x [Using product rule of differentiation] ⇒ d u = x d y + y d x 1 - x 2 y 2 Thus, from (1), we have d u = x d x Integrating both sides, we get ∫ d u = ∫ x d x ⇒ u = x 2 2 + c ⇒ sin - 1 x y = x 2 2 + c [Substitute u = sin - 1 x y ] Which is the required solution.

Solve the following differential equation : x 2 d y d x + x y + 1 - x 2 y 2 = 0

Given, x 2 d y d x + x y + 1 - x 2 y 2 = 0 ⇒ x 2 d y + x y d x = - 1 - x 2 y 2 d x ⇒ x x d y + y d x = - 1 - x 2 y 2 d x ⇒ x d y + y d x 1 - x 2 y 2 = - 1 x d x ....(1) Let u = sin - 1 x y Differentiating both the sides, d u = 1 1 - x y 2 d x y [Using chain rule of differentiation] ⇒ d u = 1 1 - x 2 y 2 x d y + y d x [Using product rule of differentiation] ⇒ d u = x d y + y d x 1 - x 2 y 2 Thus, from (1), we have d u = - 1 x d x Integrating both sides, we get ∫ d u = - ∫ 1 x d x ⇒ u = - log x + c ⇒ sin - 1 x y + log x = c [Substitute u = sin - 1 x y ] Which is the required solution.

Find the particular solution of the following differential equation: d y - d x d y + d x = x y , given that y = 1 , when x = 0 .

Given, d y - d x d y + d x = x y ⇒ y d y - y d x = x d y + x d x ⇒ y d y - x d x = x d y + y d x ⇒ y d y - x d x = d x y [Product rule : d d x x y = x d y d x + y d x d x ] Integrating both sides, we get ∫ y d y - ∫ x d x = ∫ d x y ⇒ y 2 2 - x 2 2 = x y + c Given that when x = 0 , y = 1 ⇒ 1 2 - 0 2 = 0 + c ⇒ c = 1 2 So, y 2 2 - x 2 2 = x y + 1 2 ∴ y 2 - x 2 = 2 x y + 1 Which is the required particular solution.

Find the particular solution of the following differential equation: x d x + y d y + x d y - y d x x 2 + y 2 = 0 , given that y = 1 , when x = 1 .

Given, x d x + y d y + x d y - y d x x 2 + y 2 = 0 ....(1) Let u = tan - 1 y x Differentiating both the sides, d u = 1 1 + y x 2 d y x [Using chain rule of differentiation] ⇒ d u = x 2 x 2 + y 2 x d y - y d x x 2 [Using quotient rule of differentiation] ⇒ d u = x d y - y d x x 2 + y 2 Thus, from (1), we have x d x + y d y + d u = 0 Integrating both sides, we get ∫ x d x + ∫ y d y = - ∫ d u ⇒ x 2 2 + y 2 2 = - u + c ⇒ x 2 + y 2 2 + tan - 1 y x = c [Substitute u = tan - 1 y x ] Given that when x = 1 , y = 1 ⇒ 1 + 1 2 + tan - 1 1 = c ⇒ c = 1 + π 4 So, x 2 + y 2 2 + tan - 1 y x = 1 + π 4 Which is the required particular solution.

Find the particular solution of the following differential equation : x y 2 - e 1 x 3 d x - x 2 y d y = 0 , given that y = 0 , when x = 1 .

Given, x y 2 - e 1 x 3 d x - x 2 y d y = 0 ⇒ x y 2 d x - x 2 y d y = e 1 x 3 d x ⇒ x y y d x - x d y = e 1 x 3 d x ⇒ - y x x d y - y d x x 2 = 1 x 4 e 1 x 3 d x ....(1) Let u = y x , then d u = x d y - y d x x 2 [Using Quotient rule of differentiation] Let v = 1 x 3 ⇒ d v = - 3 x 4 d x ⇒ 1 x 4 d x = - d v 3 Substituting these in Eq(1), we get - u d u = - e v d v 3 ⇒ u d u = e v d v 3 Integrating both sides, we get ∫ u d u = ∫ e v d v 3 ⇒ u 2 2 = 1 3 e v + c Substituting u = y x , v = 1 x 3 , we get 1 2 y x 2 = 1 3 e 1 x 3 + c ....(2) Given that when x = 1 , y = 0 ⇒ 1 2 0 = 1 3 e 1 + c ⇒ c = - e 3 substituting it in Eq(2), we get 1 2 y x 2 = 1 3 e 1 x 3 - e Which is the required particular solution.

Find the particular solution of the following differential equation : y sin x d y d x = cos x sin x - y 2 2 , when y = 1 at x = π 2 .

Given, y sin x d y d x = cos x sin x - y 2 2 ⇒ y sin x d y = cos x sin x d x - y 2 2 cos x d x ⇒ 2 y sin x d y + y 2 cos x d x = 2 cos x sin x d x ⇒ 2 y sin x d y + y 2 cos x d x = sin 2 x d x ....(1) Let u = y 2 sin x Differentiating both the sides, d u = y 2 d sin x + sin x d y 2 [Using product rule of differentiation] ⇒ d u = y 2 cos x d x + 2 y sin x d y [Using chain rule of differentiation] Thus, from (1), we have d u = sin 2 x d x Integrating both sides, we get ∫ d u = ∫ sin 2 x d x ⇒ u = - cos 2 x 2 + c Substitute u = y 2 sin x ⇒ y 2 sin x + cos 2 x 2 = c Given that when x = π 2 , y = 1 ⇒ 1 2 sin π 2 + cosπ 2 = c ⇒ c = 1 - 1 2 = 1 2 So, y 2 sin x + cos 2 x 2 = 1 2 ⇒ 2 y 2 sin x + cos 2 x = 1 Which is the required particular solution.

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - D. Differential Equation>Chapter 22 - First Order First Degree Differential Equation>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.