Question

Solve the following differential equation.

`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/xdx`

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_1|`

∴ log |1 + t² | = 2 log |x| + 2log |c₁|

= log |x²| + log |c|    …[logc₁² = log c]

∴ log |1 + t²| = log |cx ²|

∴ 1 + t² = cx²

∴ `1+ y^2/x^2 = cx^2`

∴ x² + y² = cx⁴