The general solution of the differential equation \( x^2\displaystyle\frac{dy}{dx}+y^2=xy \) is :(take \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{x}{y}=\log x+C \), \( C \) is an arbitrary constant
Dividing by \( x^2 \),
\( \displaystyle \frac{dy}{dx}=\frac{xy-y^2}{x^2}=\frac{y}{x}-\left(\frac{y}{x}\right)^2 \)
The right side depends only on y/x, so this is a homogeneous equation. Put \( y=vx \), \( \frac{dy}{dx}=v+x\frac{dv}{dx} \):
\( \displaystyle v+x\frac{dv}{dx}=v-v^2 \)
\( \displaystyle x\frac{dv}{dx}=-v^2 \)
\( \displaystyle -\frac{dv}{v^2}=\frac{dx}{x} \)
Integrating, since \( \int -v^{-2}\,dv=\frac{1}{v} \):
\( \displaystyle \frac{1}{v}=\log x+C \)
\( \displaystyle \frac{x}{y}=\log x+C \)
Shortcut: Differentiate Option (1): \( \frac{y-x\,y'}{y^2}=\frac{1}{x} \), so \( xy-x^2y'=y^2 \), which rearranges to \( x^2y'+y^2=xy \).