Question

`xy dy/dx  = x^2 + 2y^2`

Answer 1

`xy  dy/dx  = x^2 + 2y^2`

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

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

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

∴ `t/(1+t^2) dt = 1/x dx`

Integrating on both sides, we get

`1/2 int (2t)/(1+t^2)  dt = int dx/x`

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

∴ log |1 + t² | = 2 log |x| + 2log |c|

= log |x² | + log |c²|

∴ log |1 + t² | = log |c² x²|

∴ 1 + t² = c²x²

∴ `1 + y^2/x^2 = c^2x^2`

∴ x² + y² = c² x⁴