Question

Solve:$y\frac{dy}{dx}-\frac{4x}{3}-\frac{y^2}{3x}=0$

Answer 1

$y\frac{dy}{dx}+\left(\frac{1}{3x}\right)y^2=\frac{4x}{3}$

$2y\frac{dy}{dx}+\left(\frac{2}{3x}\right)y^2=\frac{8x}{3}$

put $y^2=u$

$2y\frac{dy}{dx}=\frac{du}{dx}$

$\frac{du}{dx}+\frac{2}{3x}u=\frac{8x}{3}$

$I.F=C^{\frac{2}{3}\int \frac{1}{x}x}=C^{log{x}^{\frac{2}{3}}}=x^{\frac{2}{3}}$

$u{x}^{2/3}=\int \frac{8x}{3}.x^{2/3}dx$

$y^2{x}^{2/3}=\frac{8}{3}\frac{x^{5/3+1}}{\frac{5}{3}+1}$

$x^{2/3}{y}^2=\frac{8}{3}\frac{x^{8/3}}{8/3}$

$\Rightarrow x^2{y}^6=x^8$

$\Rightarrow y^6=x^6$

$\Rightarrow x=y$