Question

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

\[\frac{dy}{dx} - y = \cos 2x\]

Solve the following differential equation:-

\[\frac{dy}{dx} - y = \cos 2x\]

Answer 1

We have, 
\[\frac{dy}{dx} - y = \cos 2x . . . . . (1)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = - 1\]
\[Q = \cos 2x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{- \int dx} \]
\[ = e^{- x} \]
\[\text{ Multiplying both sides of }(1)\text{ by }e^{- x} ,\text{ we get }\]
\[ e^{- x} \left( \frac{dy}{dx} - y \right) = e^{- x} \cos 2x \]
\[ \Rightarrow e^{- x} \frac{dy}{dx} - e^{- x} y = e^{- x} \cos 2x\]
Integrating both sides with respect to x, we get
\[y e^{- x} = \int e^{- x} \cos 2x dx + C \]
\[ \Rightarrow y e^{- x} = I + C . . . . . (2)\]
Where, 
\[I = \int e^{- x} \cos 2x dx . . . . . (3)\]
\[ \Rightarrow I = \frac{1}{2} e^{- x} \sin 2x - \frac{1}{2}\int\left( - e^{- x} \sin 2x \right) dx\]
\[ \Rightarrow I = \frac{1}{2} e^{- x} \sin 2x + \frac{1}{2}\int e^{- x} \sin 2x dx\]
\[ \Rightarrow I = \frac{1}{2} e^{- x} \sin 2x - \frac{1}{4} e^{- x} \cos 2x - \frac{1}{2} \times \frac{1}{2}\int\left[ \left( - e^{- x} \right) \times \left( - \cos 2x \right) \right] dx\]
\[ \Rightarrow I = \frac{1}{2} e^{- x} \sin 2x - \frac{1}{4} e^{- x} \cos 2x - \frac{1}{4}\int e^{- x} \cos 2x dx\]
\[ \Rightarrow I = \frac{1}{2} e^{- x} \sin 2x - \frac{1}{4} e^{- x} \cos 2x - \frac{1}{4}I .........\left[\text{ From (3)}\right]\]
\[ \Rightarrow \frac{5}{4}I = \frac{1}{2} e^{- x} \sin 2x - \frac{1}{4} e^{- x} \cos 2x\]
\[ \Rightarrow 5I = 2 e^{- x} \sin 2x - e^{- x} \cos 2x\]
\[ \Rightarrow I = \frac{e^{- x}}{5}\left( 2\sin 2x - \cos 2x \right) . . . . . (4)\]
From (2) and (4) we get
\[ \Rightarrow y e^{- x} = \frac{e^{- x}}{5}\left( 2\sin 2x - \cos 2x \right) + C\]
\[ \Rightarrow y = \frac{1}{5}\left( 2\sin 2x - \cos 2x \right) + C e^x \]
\[\text{ Hence, }y = \frac{1}{5}\left( 2\sin 2x - \cos 2x \right) + C e^x\text{ is the required solution.}\]