#### eNtRopY

A few days ago in the physics board, we encountered the following differential equation of the form:

y' = k * y^(1/2).

It's been a long time since I've taken a class in the techniques of solving differential equation, and I can't remember off-hand how to solve this problem. Rather than research the method for solving this problem myself, I thought I'd give you dudes a chance to impress all the beautiful ladies on these boards (e.g. Entropia, Kerrie, Galatea, Kat, Monique, Saint, etc.).

eNtRopY

P.S. If you are a lady, and I have forgotten you, I'm sorry... I don't have a very good memory for names.

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#### Hurkyl

Staff Emeritus
Gold Member
In other words, you're lazy and you want us to do your work for you.

This equation is seperable:

dy/dx = k * y^(1/2)

y^(-1/2) * dy = k * dx

#### eNtRopY

Lazy... or too busy?
Anyway,

y = (k*x/2 + C)^2

Thanks... I don't know why I didn't see that right away. I guess I should be embarrassed.

eNtRopY

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#### plus

Originally posted by eNtRopY
Lazy... or too busy?
Anyway,

y = (k*x/2)^2

Thanks... I don't know why I didn't see that right away. I guess I should be embarrassed.

eNtRopY
You forgot to add a constant before squaring.

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