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प्रश्न
Solve the following differential equation:-
\[\frac{dy}{dx} - y = \cos x\]
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उत्तर
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
\[ \therefore y e^{- x} = \left( \sin x - \cos x \right) e^{- x} + C\]
\[ \Rightarrow y = \frac{1}{2}\left( \sin x - \cos x \right) + C e^x\]
