Question

The solution of $x^2{y}_1^2+xy{y}_1-6y^2=0$ are (here $y_1=dy/dx)$ ?

Answer 1

$x^2{\left(\frac{dy}{dx}\right)}^2+xy\left(\frac{dy}{dx}\right)-b{y}^2=0$

Let $x\left(\frac{dy}{dx}\right)=v$

Re-writing the differentiate equation,

$\Rightarrow (v{)}^2+vy-b{y}^2=0\Rightarrow v^2+vy-b{y}^2=0$

factorizing the quadratic, $(v+3y)(v-2y)=0$

$v=-3y$ or $v=2y$

$\Rightarrow x\left(\frac{dy}{dx}\right)=-3y$ or $x\left(\frac{dy}{dx}\right)=2y$

$\Rightarrow \frac{dy}{y}=-3\frac{dx}{x}$ or $\frac{dy}{y}=2\frac{dx}{x}$

$\Rightarrow \int \frac{dy}{y}=-3\int \frac{dx}{x}$ or $\int \frac{dy}{y}=2\int \frac{dx}{x}$

$\Rightarrow y=\frac{c_1}{x^3}$ or $y=c_2{x}^2$

$\therefore$ We have two solutions $y=c{x}^2$ or $x^{3}y=c$