Question

The solution of $\frac{dy}{dx}=\frac{y^2}{xy-x^2}$.

• A. $y=c{e}^{xy}$
• B. $y=\frac{e^{\frac{y}{x}}}{c}$
• C. $logy=xy+c$
• D. $logx=xy+c$

Answer 1

The correct option is B $y=\frac{e^{\frac{y}{x}}}{c}$

$\frac{dy}{dx}=\frac{y^2}{xy-x^2}\frac{dy}{dx}=\frac{y^2}{x^2\left(\frac{y}{x}-1\right)}Continue\frac{dy}{dx}=\frac{{\left(\frac{y}{x}\right)}^2}{\left(\frac{y}{x}-1\right)}\to \left(i\right)\left[Itisogenouse{q}^n\right]Lety=vx,\frac{y}{x}=v\frac{dy}{dx}=v+\frac{xdv}{dx}andputine{q}^n\Rightarrow v+\frac{xdv}{dx}=\frac{v^2}{\left(v-1\right)}\Rightarrow \frac{xdv}{dx}=\frac{v^2}{v-1}-v\Rightarrow \frac{xdv}{dx}=\frac{v^2-v^2+v}{v-1}\Rightarrow \int \frac{\left(v-1\right)}{v}dx=\int \frac{dx}{x}\Rightarrow \int \frac{v}{v}dv-\int \frac{1}{v}dv=\log x+c\Rightarrow \int dv-\log v=\log x+c\Rightarrow v-\log v=\log x+c\Rightarrow \frac{y}{x}-\log \frac{y}{x}=\log xc\Rightarrow \frac{y}{x}=\log xc+\log \frac{y}{x}\Rightarrow \frac{y}{x}=\log cy\therefore cy=e^{\frac{y}{x}}y=\frac{e^{\frac{y}{x}}}{c}$