# Q4.    Prove that $x^2 - y^2 = c (x^2 + y^2 )^2$ is the general solution of differential equation$(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$, where c is a parameter.

H Harsh Kankaria

Given,

$(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$

$\implies \frac{ dy}{dx} = \frac{(x^3 - 3x y^2 )}{(y^3 - 3x^2 y)}$

Now, let y = vx

$\implies \frac{ dy}{dx} = \frac{ d(vx)}{dx} = v + x\frac{dv}{dx}$

Substituting the values of y and y' in the equation,

$v + x\frac{dv}{dx} = \frac{(x^3 - 3x (vx)^2 )}{((vx)^3 - 3x^2 (vx))}$

$\\\implies v + x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v}\\ \implies x\frac{dv}{dx} = \frac{1 - 3v^2 }{v^3 - 3v} -v = \frac{1 - v^4 }{v^3 - 3v}$

$\implies (\frac{v^3 - 3v }{1 - v^4})dv = \frac{dx}{x}$

Integrating both sides we get,

Now,

Let

$\implies$

$\implies$

Now,

Let v2 = p

Now, substituting the values of I1 and I2 in the above equation, we get,

Thus,

$\\ (x^2 - y^2)^2 = C'^4(x^2 + y^2 )^4 \\ \implies (x^2 - y^2) = C'^2(x^2 + y^2 )^2 \\ \implies (x^2 - y^2) = K(x^2 + y^2 )^2, where\ K = C'^2$

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