Help with DiffEQ / Calculus

[QueHuong]

Studying seperable equations right now, and I'm stuck on understanding one part of the technique:

how does H'(x) + G'(y)dy/dx = 0 equal to d/dx[H(x) + G(y)] = 0?

I see how H'(x) is d/dx[H(x)] but how does d/dx[G(y)] = G'(y)dy/dx? Shouldn't it be d/dx[G(y)y]?

 

[Rkonster]

You have to use the chain rule when you do d/dx[G(y)] .

 

[dullard]

Calculus operators work just like algebra (althought the math people hate it).

So this is true:
dG/dx = (dG/dy) * (dy/dx).

See how the dy parts cancel on the right (if you multiply and divide by the same thing, it cancels out)? You do not have the function G(x). Therefore you cannot solve the left part of my equation above. But you do have G(y). So you can solve the right portion of my equation above.

By the way, using primes (') in your post is very bad practice. Why? Does F'(y) refer to dF/dx or dF/dy? No one knows. It just leads to confusion. Once you go multivariable, you need to stop using the prime writing shortcut.

 

[Ausm]

I would help but I took Dif Eq 20 yeaers ago.


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