Question

x²y dx – (x³ + y³) dy = 0

Answer 1

x²y dx – (x³ + y³) dy = 0

∴  x²y dx – (x³ + y³) = dy 

∴  `dy/dx = (x^2y)/(x^3 + y^3)`  …(i)

Put y = tx  …(ii)

Differentiating w.r.t. x, we get

`dy/dx = t + x dt/dx`  …(iii)

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

`t + x dt/dx = (x^2 . tx)/(x^3 + t^3 x^3)`

∴ `t + x dt/dx = (x^3.t)/(x^3(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^4dt = - 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^-4 dt + int 1/t dt = - int1/x dx`

∴ `t^3/-3 + log |t| = - log |x| + c`

∴ `-1/(3t^3)+ log | t | = - log |x| + c`

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

∴`x^3/(3y^3) + log |y| - log |x| = - log |x| + c`

∴`log |y| - x^3/ (3y^3) = c`