# Seperable Equation

1. Sep 18, 2007

### 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
$$\int$$h(y)dy = $$\int$$g(x)dx
H(y) = G(x) + C

3. http://img354.imageshack.us/img354/6475/mathdl0.jpg [Broken]

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?

Last edited by a moderator: May 3, 2017
2. Sep 18, 2007

### Dick

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

3. Sep 19, 2007

### sourlemon

so I should be integrating

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

4. Sep 19, 2007

### dynamicsolo

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

5. Sep 19, 2007

### sourlemon

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

6. Sep 19, 2007

### dynamicsolo

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

7. Sep 19, 2007

### sourlemon

du dv right?

I think I got it!!! thank you so much!!!

8. Sep 19, 2007

### dynamicsolo

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