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**Solve the following differential equation:**

(x^{2} – y^{2})dx + 2xy dy = 0

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#### Solution

Differential equation is

(x^{2} – y^{2}) dx + 2xy dy = 0

`\implies dy/dx = (-(x^2 - y^2))/(2xy)` ...(1)

Which is a Homogeneous differential equation then put y = Vx in (1)

By (1) `d/dx (Vx) = ((Vx)^2 - x^2)/(2x(Vx))`

`\implies V + x (dV)/dx = (V^2 - 1)/(2V)`

`\implies x (dV)/dx = (V^2 - 1)/(2V) - V`

`\implies x (dV)/dx = (V^2 - 1 - 2V^2)/(2V)`

`\implies x (dV)/dx = (-V^2 - 1)/(2V)`

`\implies (2V)/(V^2 + 1) dV = - dx/x`

Now integrating both sides

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

`\implies` log (V^{2} + 1) = – log x + log c

`\implies log(V^2 + 1) = log(C/x)`

`\implies` V^{2} + 1 = `C/x`

`\implies y^2/x^2 + 1 = C/x`

`\implies` y^{2} + x^{2} = Cx.

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