Q.Solve the differential equation $\dfrac{dy}{dx} = \dfrac{y}{x}$ (for $x, y > 0$) by the method of variable separation.

Separating gives an integral leading to logarithms; combine them into a single relation.

The equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ separates directly into $\dfrac{dy}{y} = \dfrac{dx}{x}$.

Integrating both sides, using $\int\dfrac1t\,dt=\ln|t|+C$: $\ln|y| = \ln|x| + c$.

Rearranging: $\ln|y| - \ln|x| = c$, i.e. $\ln\left|\dfrac{y}{x}\right| = c$.

Exponentiating both sides: $\dfrac{y}{x} = e^c$. Since $e^c$ is itself just some positive constant, write $k=e^c$ (an arbitrary constant), giving $y = kx$. …

Alternative — recognising the result directly: $y=kx$ is precisely the one-constant family whose formation (by the method of §3) reproduces $x\dfrac{dy}{dx}=y$, i.e. $\dfrac{dy}{dx}=\dfrac{y}{x}$ — solving by …

A frequent slip is leaving the answer as $\ln|y|=\ln|x|+c$ without exponentiating to reach the cleaner algebraic relation $y=kx$ — a solution expressed with an un-combined logarithm and an additive constant is not considered fully simplifie …