Solve Seperable Equation: Step-by-Step Guide

[sourlemon]

[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
\inth(y)dy = \intg(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^{-y} = ex + -ye^{-y}. Can I say that -C = C? But what about e^{-y} . Did I inegrate it right?

Thank you in advance.

 

[Dick]

Nope. Didn't integrate it right. Try taking d/dy on the left side. What went wrong?

 

[sourlemon]

so I should be integrating

\underline{d}= \underline{d(e^{y})}
dx(e^{x}) dy (y-1)

 

[dynamicsolo]

You were OK up to the next to the last step. How do you integrate y·exp(-y) ?

 

[sourlemon]

I multiplied ye^{-y}dy - e^{-y}dy, then integrate

 

[dynamicsolo]

I multiplied ye^{-y}dy - e^{-y}dy, then integrate

Right, and you integrated the *second* term correctly. What integration technique must you use on the term ye^{-y} ?

 

[sourlemon]

du dv right?

I think I got it! thank you so much!

 

[dynamicsolo]

du dv right?

If you mean by that, "integration by parts", we are in agreement. I hope that works out for you...