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?

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

Quote:

Originally Posted by

**mathDad** 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?

; multiply both sides by x

; substitute in implicit function definition

1 Attachment(s)

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