Question

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

Answer 1

We have,
\[\left( x^2 - 1 \right)\frac{dy}{dx} + 2\left( x + 2 \right)y = 2\left( x + 1 \right)\]
\[ \Rightarrow \frac{dy}{dx} + \frac{2\left( x + 2 \right)}{x^2 - 1}y = \frac{2\left( x + 1 \right)}{x^2 - 1} \]
\[ \Rightarrow \frac{dy}{dx} + \frac{2\left( x + 2 \right)}{x^2 - 1}y = \frac{2}{x - 1} . . . . . . . . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = \frac{2\left( x + 2 \right)}{x^2 - 1}\]
\[Q = \frac{2}{x - 1}\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\frac{2\left( x + 2 \right)}{x^2 - 1} dx} \]
\[ = e^{\int\frac{2x}{x^2 - 1} + 4\int \frac{1}{x^2 - 1}dx} \]
\[ = e^{log\left| x^2 - 1 \right| + 4 \times \frac{1}{2}\log\left| \frac{x - 1}{x + 1} \right|} \]
\[ = e^{log \left| \left( x^2 - 1 \right) \times \frac{\left( x - 1 \right)^2}{\left( x + 1 \right)^2} \right|} \]
\[ = e^{log\left| \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)} \right|} \]
\[ = \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }\frac{\left( x - 1 \right)^3}{\left( x + 1 \right)},\text{ we get }\]
\[\frac{\left( x - 1 \right)^3}{\left( x + 1 \right)} \left( \frac{dy}{dx} + \frac{2\left( x + 2 \right)}{x^2 - 1}y \right) = \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)} \times \frac{2}{x - 1}\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}\frac{dy}{dx} - \frac{2\left( x + 2 \right) \left( x - 1 \right)^2}{\left( x + 1 \right)^2}y = \frac{2 \left( x - 1 \right)^2}{\left( x + 1 \right)}\]
Integrating both sides with respect to x, we get
\[\frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int\frac{2 \left( x - 1 \right)^2}{\left( x + 1 \right)} dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int2\left\{ \frac{\left( x + 1 \right)^2 - 4x}{\left( x + 1 \right)} \right\} dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int2\left\{ \left( x + 1 \right) - \frac{4x}{\left( x + 1 \right)} \right\} dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int2\left\{ \left( x + 1 \right) - \frac{4\left( x + 1 - 1 \right)}{\left( x + 1 \right)} \right\} dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int2\left\{ \left( x + 1 \right) - 4 + \frac{4}{\left( x + 1 \right)} \right\} dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = \int\left( 2x - 6 + \frac{8}{x + 1} \right) dx + C\]
\[ \Rightarrow \frac{\left( x - 1 \right)^3}{\left( x + 1 \right)}y = x^2 - 6x + 8\log \left| x + 1 \right| + C\]
\[ \Rightarrow y = \frac{\left( x + 1 \right)}{\left( x - 1 \right)^3}\left( x^2 - 6x + 8\log \left| x + 1 \right| + C \right)\]
\[\text{Hence, }y = \frac{\left( x + 1 \right)}{\left( x - 1 \right)^3}\left( x^2 - 6x + 8\log\left| x + 1 \right| + C \right)\text{ is the required solution.} \]