Question

Solve the following differential equation:-

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

Answer 1

We have,

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

\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]

\[P = - 1 \]

\[Q = \cos x\]

Now,

\[ I . F . = e^{- 1\int dx} = e^{- x} \]

Solution is given by,

\[y \times I . F . = \int\cos x \times I . F . dx + C\]

\[ \Rightarrow y e^{- x} = \int e^{- x} \cos x dx + C\]

\[ \Rightarrow y e^{- x} = I + C . . . . . \left( 1 \right)\]

Where,

\[ \Rightarrow I = \cos x\int e^{- x} dx - \int\left[ \frac{d}{dx}\left( \cos x \right)\int e^{- x} dx \right]dx\]

\[ \Rightarrow I = - \cos x e^{- x} - \int\sin x e^{- x} dx\]

\[ \Rightarrow I = - \cos x e^{- x} - \sin x\int e^{- x} dx + \int\left[ \frac{d}{dx}\left( \sin x \right)\int e^{- x} dx \right]dx\]

\[ \Rightarrow I = - \cos x e^{- x} + \sin x e^{- x} - \int\left[ \cos x e^{- x} \right]dx\]

\[ \Rightarrow I = - \cos x e^{- x} + \sin x e^{- x} - I ..........\left[\text{Using (2)} \right]\]

\[ \Rightarrow 2I = - \cos x e^{- x} + \sin x e^{- x} \]

\[ \Rightarrow I = \frac{1}{2}\left( - \cos x + \sin x \right) e^{- x} . . . . . . . . \left( 3 \right)\]

From (1) and (3), we get

\[ \therefore y e^{- x} = \left( \sin x - \cos x \right) e^{- x} + C\]

\[ \Rightarrow y = \frac{1}{2}\left( \sin x - \cos x \right) + C e^x\]