Answer to Question #188113 in Differential Equations for ernest

Question

y2 dx — x(2x + 3y)dy = 0

Answer 1

We have given the differential equation,

$y^2dx-x(2x+3y)dy = 0$

$y^2dx = x(2x+3y)dy$

$\dfrac{dy}{dx} = \dfrac{y^2}{x(2x+3y)}$

Putting $y = vx$

Hence, $\dfrac{dy}{dx} = v+ x \dfrac{dv}{dx}$

Therefore,

$v+ x \dfrac{dv}{dx} = \dfrac{v^2x^2}{x(2x+3vx)}$

$x \dfrac{dv}{dx} = \dfrac{-2v^2-2v}{3v+2}$

$x \dfrac{dv}{dx} = \dfrac{-v(2v+2)}{3v+2}$

$-\int \dfrac{(3v+2)dv}{2v(v+1)} = \int \dfrac{dx}{x}$

$-\dfrac{ln(v+1)}{2} - lnv = lnx + lnC$

Replacing $v = \dfrac{y}{x}$

We get,

$-\dfrac{ln(\dfrac{y}{x}+1)}{2} - ln\dfrac{y}{x} = lnx + lnC$

This is the required solution of given differential equation.