Solve Seperable Equation: Step-by-Step Guide

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sourlemon
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[SOLVED] Seperable Equation

1. Instruction: Solve the equation.

2. Equations:
dy/dx = g(x)p(y)
h(y) = 1/p(y)
h(y)dy = g(x)dx
[tex]\int[/tex]h(y)dy = [tex]\int[/tex]g(x)dx
H(y) = G(x) + C


3. http://img354.imageshack.us/img354/6475/mathdl0.jpg

I tried to do it on the right side, but...I got stuck there. If I add to the right, then I would be left with -C + e[tex]^{-y}[/tex] = ex + -ye[tex]^{-y}[/tex]. Can I say that -C = C? But what about e[tex]^{-y}[/tex] . Did I inegrate it right?

Thank you in advance.
 
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so I should be integrating

[tex]\underline{d}[/tex]= [tex]\underline{d(e^{y})}[/tex]
dx(e[tex]^{x}[/tex]) dy (y-1)
 
I multiplied ye[tex]^{-y}[/tex]dy - e[tex]^{-y}[/tex]dy, then integrate
 
sourlemon said:
I multiplied ye[tex]^{-y}[/tex]dy - e[tex]^{-y}[/tex]dy, then integrate

Right, and you integrated the *second* term correctly. What integration technique must you use on the term ye[tex]^{-y}[/tex] ?
 
du dv right?

I think I got it! thank you so much!