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^2y)dy, where c is a parameter.

Here, x^2-y^2=c(x^2+y^2)^2 ... (1) Differentiating w.r.t.x, we get, 2x-2y dy dx =2c(x^2+y^2). [2x+2y dy dx ] x-y dy dx =2c(x^2+y^2). [x+y dy dx ] ...(2) Dividing (2) by (1), we get. x-y dy dx x^2-y^2 = 2 (x+y dy dx ) x^2+y^2 x(x^2+y^2)-y(x^2+y^2) dy dx =2x(x^2-y^2)+2y(x^2-y^2) dy dx [2y(x^2-y^2)+y(x^2+y^2) ] dy dx =x(x^2+y^2)-2x(x^2-y^2) (2x^2y-2y^3+x^2y+y^3) dy dx =x^3+xy^2-2x^3+2xy^2 (3x^2y-y^3) dy dx =3xy^2-x^3 dy dx = x^3-3xy^2 y^3-3x^2y which is the required equation. Hence the result.

Prove that x^2-y^2=c(x^2+y^2)^2 is the general solution of differ…

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

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Answer

$\text{Here},\ \ \text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2\ \ ...{(1)}$
Differentiating w.r.t.x, we get,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]$
$\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big]\ \ ...(2)$
Dividing (2) by (1), we get.
$\frac{\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}}{\text{x}^2-\text{y}^2}=\frac{2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)}{\text{x}^2+\text{y}^2}$
$\therefore\ \ \text{x}(\text{x}^2+\text{y}^2)-\text{y}(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}$ $=2\text{x}(\text{x}^2-\text{y}^2)+2\text{y}(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}$
$\therefore\ \ \big[2\text{y}(\text{x}^2-\text{y}^2)+\text{y}(\text{x}^2+\text{y}^2)\big]\frac{\text{dy}}{\text{dx}}$ $=\text{x}(\text{x}^2+\text{y}^2)-2\text{x}(\text{x}^2-\text{y}^2)$
$\therefore\ \ (2\text{x}^2\text{y}-2\text{y}^3+\text{x}^2\text{y}+\text{y}^3)\frac{\text{dy}}{\text{dx}}$ $=\text{x}^3+\text{xy}^2-2\text{x}^3+2\text{xy}^2$
$\therefore\ \ (3\text{x}^2\text{y}-\text{y}^3)\frac{\text{dy}}{\text{dx}}=3\text{xy}^2-\text{x}^3$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^3-3\text{xy}^2}{\text{y}^3-3\text{x}^2\text{y}}$ which is the required equation.
Hence the result.

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