Thread: y' = (x^2 + 3y^2) / (2xy)

1. y' = (x^2 + 3y^2) / (2xy)

In a DiffEQ video, the problem y' = (x^2 + 3y^2) / (2xy) is solved as a homogeneous differential equation. The steps are all fine and the solution is found to be x^2 + y^2 - Cx^3 = 0.

If the answer is correct, then the derivative of the solution should equal the RHS of the question. Am I understanding DiffEQs correctly?

If yes, then what's wrong here? I take the derivative of the solution and solve for y'
2x + 2y y' -3Cx^2 = 0
y' = (3Cx^2 - 2x) / (2y).

But I don't see how to make that look like the question. My derivative looks correct, so what am I missing?

2. Re: y' = (x^2 + 3y^2) / (2xy)

In a DiffEQ video, the problem y' = (x^2 + 3y^2) / (2xy) is solved as a homogeneous differential equation. The steps are all fine and the solution is found to be x^2 + y^2 - Cx^3 = 0.

If the answer is correct, then the derivative of the solution should equal the RHS of the question. Am I understanding DiffEQs correctly?

If yes, then what's wrong here? I take the derivative of the solution and solve for y'
2x + 2y y' -3Cx^2 = 0
y' = (3Cx^2 - 2x) / (2y).

But I don't see how to make that look like the question. My derivative looks correct, so what am I missing?
$\displaystyle 2x+2y y' - 3Cx^2=0$ ; multiply both sides by x

$\displaystyle 2x^2 +2x y y' = 3C x^3$

$\displaystyle 2x y y'=3Cx^3 - 2x^2$ ; substitute in implicit function definition

$\displaystyle 2x y y'=3(x^2 + y^2) - 2x^2$

$\displaystyle 2x y y'=x^2+3y^2$

$\displaystyle y' = \frac{x^2+3y^2}{2x y}$