Sum

Solve the following differential equation:-

\[\left( 1 + x^2 \right)\frac{dy}{dx} - 2xy = \left( x^2 + 2 \right)\left( x^2 + 1 \right)\]

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#### Solution

Given,

\[\left( 1 + x^2 \right)\frac{dy}{dx} - 2xy = \left( x^2 + 2 \right)\left( x^2 + 1 \right)\]

\[\Rightarrow \frac{dy}{dx} - \frac{2x}{\left( 1 + x^2 \right)}y = \left( x^2 + 2 \right)\]

\[\]

This is a linear differential equation

\[I.F.= e^{\int - \frac{2x}{1 + x^2}dx} = \frac{1}{1 + x^2}\]

\[y\left( \frac{1}{1 + x^2} \right) = \int\left( \frac{x^2 + 2}{x^2 + 1} \right)dx\]

\[ \Rightarrow y\left( \frac{1}{1 + x^2} \right) = \int\left( 1 + \frac{1}{1 + x^2} \right)dx\]

\[ \Rightarrow y\left( \frac{1}{1 + x^2} \right) = x + \tan^{- 1} x + C\]

\[ \Rightarrow y = \left( x + \tan^{- 1} x + C \right)\left( 1 + x^2 \right)\]

\[ \Rightarrow y\left( \frac{1}{1 + x^2} \right) = \int\left( 1 + \frac{1}{1 + x^2} \right)dx\]

\[ \Rightarrow y\left( \frac{1}{1 + x^2} \right) = x + \tan^{- 1} x + C\]

\[ \Rightarrow y = \left( x + \tan^{- 1} x + C \right)\left( 1 + x^2 \right)\]

Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations

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