Question

\[\frac{dy}{dx} + y = \cos x\]

Answer 1

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