Question

Solve the following differential equation:

(x² − y²)dx + 2xy dy = 0

Answer 1

Given:

(x² − y²)dx + 2xy dy = 0

2xy dy = −(x² − y²) dx 

∴ `"dy"/"dx" = − ("x"^2 − "y"^2)/"2xy" = (y^2 − x^2)/(2xy)`  ....[1]

Put y = vx

∴ y = vx ⇒`"dy"/"dx" = "v"+ x("dv")/"dx"`

`v + x(dv)/dx = (v^2 − 1)/(2v)`

∴ `v + x (dv)/dx = ((vx)^2 − x^2)/(2x (vx)) = (x^2(v^2 − 1))/(2vx^2) = (v^2 − 1)/(2v)`

∴ `x (dv)/dx = (v^2 − 1)/(2v) − v = (v^2 − 1 − 2v^2)/(2v) = (− (v^2 + 1))/(2v)`

⇒ `x (dv)/dx = − (v^2 + 1)/(2v)`

⇒ `(2v)/(v^2 + 1)dv = − 1/x dx`

Integrating both sides, we get:

⇒ `∫ (2v)/(v^2 + 1)dv = − ∫ 1/x dx`

⇒ log (v² + 1) = − log x + log c₁

⇒ log (x(v² + 1)) = log c₁

⇒ x(v² + 1) = c₁

⇒ `x (1 + y^2/x^2) = c_1`      ...[putting `v = y/x`]

⇒ `x + y^2/x^2 = c_1`     ...[Multiply by x]

⇒ x² + y² = c₁​x    ...[Let c = c₁]

= x² + y² = cx