Question

Solve the following differential equation.

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^2x^2)/(x.tx+x^2)` 

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

∴  `t + x dt/dx  = (-t^2x^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 = -int1/xdx`

∴`int (2t + 1 - t)/(t(2t+1)) dt = - int1/xdx`

∴`int1/tdt-int  1/ (2t+1) dt = -int1/ 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 |y² x| = log | c² (x + 2y)|

∴ xy 2 = c² (x + 2y)