For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$

Question

For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.

I solved for $\frac{dy}{dx}$ and got $$\frac{dy}{dx}=\frac{cx-3x(x^2+2y^2)^2} {6y(x^2+2y^2)^2}$$ I was told to solve for $c$ and multiply by $dx$. but I cannot figure out how to get this to the original equation. My teacher said to use implicit differentiation so I solved for $\frac{dy}{dx}$.

Answer 1

The hint says you should solve for $c$ before differentiating, so let's do that

$$ \frac{(x^2+2y^2)^3}{x^2} = c $$

Performing implicit differentiation on this form will eliminate $c$

$$ \frac{3x^2(x^2+2y^2)^2\left(2x + 4y\frac{dy}{dx}\right)-2x(x^2+2y^2)^3}{x^4} = 0 $$

Now you can solve for $\frac{dy}{dx}$ and finish the job

Answer 2

Calling

$$ f(x,y) = (x^2 + 2 y^2)^3 -c x^2 = 0 $$

we have

$$ df=f_x dx+f_y dy = 0\to y' = -\frac{f_x}{f_y} = \frac{x \left(c-3 \left(x^2+2 y^2\right)^2\right)}{6 y \left(x^2+2 y^2\right)^2} $$

now substituting $c = \frac{\left(x^2+2 y^2\right)^3}{x^2}$ and simplifying we obtain

$$ y' = \frac{y}{3 x}-\frac{x}{3 y} $$

as expected.

NOTE

now making $z = \frac yx \to y' = x z' + z$ we have the transformed DE

$$ x z' = \frac 13\left(z-\frac 1z\right)-z $$

which is separable and can be easily solved.