Question

\[\frac{dy}{dx} - y = x e^x\]

Answer 1

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