Question

y2 dx + (xy + x²)dy = 0

Answer 1

y2 dx + (xy + x²)dy = 0

∴ (xy + x² ) dy = -y² dx

∴ `dy/dx = -y^2/(xy + x^2)` ...(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 = (-t^2 x^2)/(x.tx + x^2)`

∴ `t + x dt/dx = (-t^2 x^2)/(x^2(t+1)`

∴ `x dt/dx = (-t^2)/(t+1) -t`

∴  `x dt/dx = (-t^2 - t^2 - t)/(t+1)`

∴  `x dt/dx = (- (2t^2 + t))/(t+1)`

∴ `(t+1)/(2t^2 +t) dt = -1/x dx`

Integrating on both sides, we get

`int (t+1)/(2t^2 + t) dt = - int 1/x dx`

∴ `int (2t +1 - t)/(t(2t+1)) dt = - int 1/x dx`

∴ `int 1/t dt - int 1/(2t + 1) dt = -int 1/x dx`

∴ `log | t | -1/ 2 log |2t + 1| = -log |x| + log |c|`

∴ 2log| t | -log |2t + 1| = -2log |x| + 2 log |c|

∴ `2log |y/x| -log |(2y)/ x +1|=- 2log |x| + 2 log |c|`

∴  2log |y| - 2log |x| - log |2y + x| + log |x| = - 2log |x| + 2log |c|

∴  log |y²| + log |x| = log |c² |+ log |2y + x|

∴  log |y²x| = log |c²(x + 2y)|

∴  xy² = c² (x + 2y)