Question

\[\frac{dy}{dx} + \frac{4x}{x^2 + 1}y + \frac{1}{\left( x^2 + 1 \right)^2} = 0\]

Answer 1

We have, 
\[\frac{dy}{dx} + \frac{4x}{x^2 + 1}y + \frac{1}{\left( x^2 + 1 \right)^2} = 0 \]
\[ \Rightarrow \frac{dy}{dx} + \frac{4x}{x^2 + 1}y = - \frac{1}{\left( x^2 + 1 \right)^2} . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = \frac{4x}{x^2 + 1} \]
\[Q = - \frac{1}{\left( x^2 + 1 \right)^2}\]
\[ \therefore \text{I.F.} = e^{\int P\ dx} \]
\[ = e^{2\int\frac{2x}{x^2 + 1} dx} \]
\[ = e^{2\log \left| x^2 + 1 \right|} \]
\[ = \left( x^2 + 1 \right)^2 \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }\left( x^2 + 1 \right)^2 ,\text{ we get }\]
\[ \left( x^2 + 1 \right)^2 \left( \frac{dy}{dx} + \frac{4x}{x^2 + 1}y \right) = \left( x^2 + 1 \right)^2 \left[ - \frac{1}{\left( x^2 + 1 \right)^2} \right] \]
\[ \Rightarrow \left( x^2 + 1 \right)^2 \frac{dy}{dx} + 4x\left( x^2 + 1 \right)y = - 1\]
Integrating both sides with respect to x, we get
\[ \left( x^2 + 1 \right)^2 y = - \int dx + C\]
\[ \Rightarrow \left( x^2 + 1 \right)^2 y = - x + C\]
\[\text{ Hence, }\left( x^2 + 1 \right)^2 y = - x + C\text{ is the required solution.}\]