Question

\[\frac{dy}{dx} = x e^x - \frac{5}{2} + \cos^2 x\]

Answer 1

We have,
\[\frac{dy}{dx} = x\ e^x - \frac{5}{2} + \cos^2 x\]
\[ \Rightarrow \frac{dy}{dx} = x\ e^x - \frac{5}{2} + \frac{\cos 2x}{2} + \frac{1}{2}\]
\[ \Rightarrow \frac{dy}{dx} = x\ e^x + \frac{\cos 2x}{2} - 2\]
\[ \Rightarrow dy = \left( x\ e^x + \frac{\cos 2x}{2} - 2 \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( x\ e^x + \frac{\cos 2x}{2} - 2 \right)dx\]
\[ \Rightarrow y = \int x\ e^x dx + \frac{1}{2}\int\cos 2x dx - 2\int dx\]
\[ \Rightarrow y = x\int e^x dx - \int\left[ \frac{d}{dx}\left( x \right)\int e^x dx \right]dx + \frac{1}{2} \times \frac{\sin 2x}{2} - 2x\]
\[ \Rightarrow y = x\ e^x - e^x + \frac{1}{4}\sin 2x - 2x + C\]
\[\text{ Hence, }y = x\ e^x - e^x + \frac{1}{4}\sin 2x - 2x +\text{ C is the solution to the given differential equation.}\]