Question

Find the particular solution of the differential equation:

(x² – 2y²) dx + 2xy dy = 0, when x = 1 and y = 1

Answer 1

Given differential equation is (x² – 2y²) dx + 2xy dy = 0.

`dy/dx = (-(x^2 - 2y^2))/(2xy)`

`dy/dx = (2y^2 - x^2)/(2xy)`   ...(i)

It is a homogeneous differential equation.

Put y = vx

Differentiate w.r.t. x

`dy/dx = v + x (dv)/dx`

Now, equation (i) becomes

`v + x (dv)/dx = (2v^2x^2 - x^2)/(2x xx vx)`

⇒ `v + x (dv)/dx = (2v^2 - 1)/(2v)`

⇒ `v + x (dv)/dx = v - 1/(2v)`

⇒ `x (dv)/dx = -1/(2v)`

⇒ `x  dv = - 1/(2v) dx`

⇒ `int 2v  dv = -int 1/x dx`

⇒ `(2v^2)/2 = -log x + C`

⇒ v² = – log x + C

But `v = y/x`

`y^2/x^2 = -log x + C`   ...(ii)

When x = 1 and y = 1

`1/1 = -log 1 + C`

1 = 0 + C

⇒ C = 1

Put the value of C in equation (ii)

`y^2/x^2 = -log x + 1`

⇒ y² = – x² log x + x²

So, y² + x² log x = x².