Question

Solve the following differential equations:

x²ydx – (x³ – y³)dy = 0

Answer 1

x²ydx – (x³ – y³)dy = 0

∴ x²ydx = (x³ + y³ )dy

∴ ${\displaystyle \frac{dy}{dx}=\frac{x^{2}y}{x^3+y^3}}$  ...(i)

Put y = tx  ...(ii)

Differentiating w.r.t x,

∴ ${\displaystyle \frac{dy}{dx}=t+x\frac{dt}{dx}}$ ...(iii)

Substituting (iii) and (ii) in (i), we get

∴ ${\displaystyle t+x\frac{dt}{dx}=\frac{t{x}^3}{x^3+t^3{x}^3}}$

∴ ${\displaystyle t+x\frac{dt}{dx}=\frac{t}{1+t^3}}$

∴ ${\displaystyle \frac{x.dt}{dx}=\frac{t}{1+t^3}-t}$

∴ ${\displaystyle \frac{x.dt}{dx}=\frac{t-t-t^4}{1+t^3}}$

∴ ${\displaystyle \frac{x.dt}{dx}=\frac{-t^4}{1+t^3}}$

∴ ${\displaystyle \frac{1+t^3}{t^4}.dt=-\frac{dx}{x}}$

Integrating on both sides. we get

${\displaystyle \int \frac{1+t^3}{t^4}dt=-\int \frac{1}{x}dx}$

∴ ${\displaystyle \int (\frac{1}{t^4}+\frac{1}{t})dt=-\int \frac{1}{x}dx}$

∴ ${\displaystyle \int t^{-4}dt+\int \frac{1}{t}dt=-\int \frac{1}{x}dx}$

∴  ${\displaystyle \frac{t^{-3}}{-3}+\log \left|t\right|}$ = –log |x| + log |c₁|

∴ ${\displaystyle -\frac{1}{3t^3}+\log \left|t\right|}$ = –log |x| + log |c₁|

∴ ${\displaystyle \frac{-1}{+3}.\frac{1}{{\left(\frac{y}{x}\right)}^3}+\log \left|\frac{y}{x}\right|}$ = –log |x| + log |c₁|

∴ ${\displaystyle -\frac{x^3}{3y^3}+\log \text{|y| - log}\left|x\right|}$ = –log |x| + log |c₁| 

∴ log |y| + log |c| = ${\displaystyle \frac{x^3}{3y^3}}$

Where [–log |c₁| = log |c|]

∴ log |yc| = ${\displaystyle \frac{x^3}{3y^3}}$

This is the general solution.