Question

Solve the following differential equation:

x² dy + y(x + y) dx = 0

Answer 1

We have, `x^2dy + y(x + y)dx = 0`

`x^2/y dy/dx + (x + y) = 0`

`x^2/y dy/dx + x/y + 1 = 0`        ...(i)

Put y = vx

⇒ `(dy)/(dx) = V + x(dV)/(dx)`

From eq (i), we get

`1/(V^2)(V + x(dV)/(dx)) + 1/V + 1 = 0`

⇒ `x(dV)/(dx) = -(2V + V^2)`

⇒ `1/2[1/V - 1/(V + 2)]dV = -(dx)/x`

On integrating both sides, we get

⇒ `1/2(log V - log V + 2) = -log x + C_1`

⇒ `1/2 log(V/(V + 2))= -log x + C_1`

⇒ `log (V/(V + 2)) = -2 log x + log k`

where `k = e^(2C_1)`

⇒ `log(V/(V + 2))= log(k/x^2)`

⇒ `V/(V + 2) = k/x^2`

Substituting V = `y/x`, we get

`y/(2x + y) = k/x^2`

⇒ `x^2y = C^2(2x + y)  "where"  k = C^2`