Question

Solve the following initial value problem:-

\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]

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 
\[\frac{dy}{dx} + Py = Q\]
\[\text{where }P = - \frac{1}{x}\text{ and }Q = \frac{x + 1}{x} e^{- x} \]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{- \int\frac{1}{x} dx} \]
\[ = e^{- \log x} \]
\[ = \frac{1}{x}\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = \frac{1}{x},\text{ we get }\]
\[\frac{1}{x} \left( \frac{dy}{dx} - \frac{1}{x}y \right) = \frac{1}{x}\left( \frac{x + 1}{x} \right) e^{- x} \]
\[ \Rightarrow \frac{1}{x} \left( \frac{dy}{dx} - \frac{1}{x}y \right) = \left( \frac{x + 1}{x^2} \right) e^{- x} \]
Integrating both sides with respect to x, we get
\[\frac{1}{x}y = \int\left( \frac{1}{x} + \frac{1}{x^2} \right) e^{- x} dx + C\]
\[\text{ Putting }\frac{1}{x} e^{- x} = t\]
\[ \Rightarrow \left( - \frac{1}{x} e^{- x} - \frac{1}{x^2} e^{- x} \right)dx = dt\]
\[ \Rightarrow \left( \frac{1}{x} + \frac{1}{x^2} \right) e^{- x} dx = - dt\]
\[ \therefore \frac{1}{x}y = \int - dt + C\]
\[ \Rightarrow \frac{y}{x} = - t + C\]
\[ \Rightarrow \frac{y}{x} = - \frac{e^{- x}}{x} + C\]
\[ \Rightarrow y = - e^{- x} + Cx . . . . . \left( 2 \right)\]
Now,
\[y\left( 1 \right) = 0\]
\[ \therefore 0 = - e^{- 1} + C\]
\[ \Rightarrow C = e^{- 1} \]
\[\text{Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y = - e^{- x} + x e^{- 1} \]
\[ \Rightarrow y = x e^{- 1} - e^{- x} \]
\[\text{Hence, }y = x e^{- 1} - e^{- x}\text{ is the required solution.}\]