Question

\[x\frac{dy}{dx} = x + y\]

Answer 1

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