Question

Find one-parameter families of solution curves of the following differential equation:-

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

Solve the following differential equation:-

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

Answer 1

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