Question

solve x(x-y)dy+y^2dx=0​

Answer 1

Answer:

The solution of the differential equation is

$\bf\frac{y}{x}=log\,y+c$

Step-by-step explanation:

I have applied variable separable method to find the solution of the given differential equation.

In this method, the variables in the given differential equation are separated in such a way that suitable for integration.

$x(x-y)dy+y^2\,dx=0$

$x^2\,dy-xy\,dy+y^2\,dx=0$

$x^2\,dy-y(x\,dy-y\,dx)=0$

$-y(x\,dy-y\,dx)=-x^2\,dy$

$\frac{x\,dy-y\,dx}{x^2}=\frac{dy}{y}$

Using quotient rule of differentiation,

$\boxed{\bf\frac{d(\frac{u}{v})}{dx}=\frac{v\,\frac{du}{dx}-u\,\frac{dv}{dx}}{v^2}}$

$d(\frac{y}{x})=\frac{dy}{y}$

Integrating on both sides, we get

$\int{d(\frac{y}{x})}=\int{\frac{dy}{y}}$

$\implies\bf\frac{y}{x}=log\,y+c$