Question

Find the general solution of the differential equation \[\frac{dy}{dx} - y = \cos x\]

Answer 1

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