I'm doing ODE review before I start PDE this next semester and appear to have forgotten everything. Problem on implicit solution/"show that" below in a moment.

Question

mendicant_bias · Answer

$$\text{Show that} \ x^2+y^2-4=0 \ \text{is an implicit solution of the first order ODE}$$$$(y')^2+1=4y^{-2}, \ \ \ (-2<x<2).$$

mendicant_bias · Answer

I remember pretty easily how to solve the explicit given solution variant where you just have y equalling some quantity and can repeatedly derive it and plug into the original ODE, but I don't think I ever show this for an implicitly given solution.

mendicant_bias · Answer

Is there any particular way I should be doing this different, or is it essentially the same?

perl · Answer

yes it sounds the same. take the implicit derivative of y(t), your given solution, and plug into the differential equation. if it satisfies it, you have shown it is an implicit solution

perl · Answer

x^2 + y^2 -4 = 0

2x + 2y * y' = 0 .   

Therefore 
y' = -x/y

mendicant_bias · Answer

Yeah, got it; thank you!

mendicant_bias · Answer

I'm saying this right now because I want to take a shot at it myself, but I'm confused at the moment about the algebra that follows, primarily how that four appears. Going to still try to figure that out by myself for a moment.

perl · Answer

sure

mendicant_bias · Answer

Nevermind, got it; subbing in the original solution value, x^2+y^2=4, for that last part. Thanks.