Question

The equation x2ydx-(x3+y3)dy= 0 is homogeneous equation

Answer 1

The given homogeneous differential equation is ....

$x^2ydx-(x^3+y^3)dy=0\\\\x^2ydx=(x^3+y^3)dy\\\\\frac{x^2y}{x^3+y^3}=\frac{dy}{dx}\\\\\frac{\frac{y}{x}}{1+\frac{y^3}{x^3}}=\frac{dy}{dx}$

if we assume y/x = t
y = xt
dy/dx = t + xdt/dx

then, t/(1 + t³) = t + x.dt/dx

t/(1 + t³) - t = xdt/dx

(t - t - t⁴)/(1 + t³) = x dt/dx

-t⁴/(1 + t³) = x dt/dx

dx/x = -(1 + t³)/t⁴ dt

lnx = -t^-4 dt - dt/t

lnx = -t^-3/-3 - lnt + C

lnx = 1/3t³ - lnt + C

now, put t = y/x

so, lnx = 1/3(y/x)³ - lny/x + C

or, lnx = x³/3y³ - lny + lnx + C

hence, -x³/3y³ + lny + C = 0