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Find the particular solution of the differential equation x(1

Question

Find the particular solution of the differential equation $x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0$ , given that $y = 1$ when $x = 0$ .

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Find the particular solution of the differential equation

x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0

, given that

y = 1

when

x = 0

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

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

The given equation is in the form of first order separable ODE has the form of

N (y) d y = M (x) d x

Let

y

be the dependent variable.

Divide by

d x

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

Rewrite in the form of first order separable ODE:

N (y) = \frac{y}{y ^{2} + 1}, M (x) = \frac{x}{x ^{2} + 1}

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

Integrating on both the sides,

\int \frac{y}{y ^{2} + 1} d y = \int \frac{x}{x ^{2} + 1} d x

\frac{1}{2} ln (y^{2} + 1) = \frac{1}{2} ln (x^{2} + 1) + C

--- (1)

When

y = 1, x = 0

Equtaion (1) becomes,

\frac{1}{2} ln (1^{2} + 1) = \frac{1}{2} ln (0^{2} + 1) + C \Rightarrow c = \frac{1}{2} ln 2

The required equation is

\frac{1}{2} ln (y^{2} + 1) = \frac{1}{2} ln (x^{2} + 1) + \frac{1}{2} ln 2.

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Question Text	Find the particular solution of the differential equation $x (1 + y^{2}) d x - y (1 + x^{2}) d y = 0$ , given that $y = 1$ when $x = 0$ .
Answer Type	Text solution:1
Upvotes	150