Question

Solve the following differential equation.

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^3x^3)`

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

∴ `x dt/dx =  t /(1+ t ^3)-t`

∴ `xdt/dx =( t-t-t^4)/(1+t^3)`

∴ `xdt/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 = - int 1/x dx`

∴ `int(1/t^4+1/t)dt = - int1/x dx`

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

∴ ` t^-3/-3+ log |t| = - log |x| + log|c_1|`

∴ ` - 1/-3t^3+ log |t| = - log |x| + log|c_1|`

∴ `-1/3. 1/(y/x)^3+log|y/x| = - log|x| + log|c_1|`

∴ `-x^3/(3y^3)+ log|y|-log|x| = -log|x| + log|c_1|`

∴ `log |y| + log |c| = x^3/(3y^3)`           ...[-log |c₁| = log |c|]

∴ `log |yc| = x^3/(3y^3)`