Question

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

\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x}\]

Solve the following differential equation:-

\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x}\]

Answer 1

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