Question

Solve the following differential equations:

x²ydx – (x³ – y³)dy = 0

Answer 1

x²ydx – (x³ – y³)dy = 0

∴ x²ydx = (x³ + y³ )dy

∴ `dy/dx = (x^2y)/(x^3 + y^3)`  ...(i)

Put y = tx  ...(ii)

Differentiating w.r.t x,

∴ `dy/dx = t + x dt/dx` ...(iii)

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

∴ `t + x dt/dx = (tx^3)/(x^3 + t^3x^3)`

∴ `t + x dt/dx = t/(1 + t^3)`

∴ `(x.dt)/dx = t/(1 + t^3) - t`

∴ `(x.dt)/dx = (t - t - t^4)/( 1 + t^3)`

∴ `(x.dt)/dx = (-t^4)/(1 + t^3)`

∴ `(1 + t^3)/t^4.dt = - dx/x`

Integrating on both sides. we get

`int (1 + t^3)/t^4 dt = - int 1/x dx`

∴ `int (1/t^4 + 1/t)dt = -int 1/x dx`

∴ `int t^-4dt + int 1/t dt = - int 1/x dx`

∴  `t^-3/(-3) + log |t|` = –log |x| + log |c₁|

∴ `- 1/(3t^3) + log""|t|` = –log |x| + log |c₁|

∴ `(-1)/(+3). 1/(y/x)^3 + log"|y/x|` = –log |x| + log |c₁|

∴ `- x^3/(3y^3) + log"|y| - log"|x|` = –log |x| + log |c₁| 

∴ log |y| + log |c| = `x^3/(3y^3)`

Where [–log |c₁| = log |c|]

∴ log |yc| = `x^3/(3y^3)`

This is the general solution.