# Homework Help: Find the particular solution of a differential equation

1. Jan 30, 2012

### TrueStar

1. The problem statement, all variables and given/known data

Find the particular solution of the differential equation dy/dx = (x-4)e^(-2y)

Satisfying initial condition y(4)=ln(4)

2. Relevant equations

N/A

3. The attempt at a solution

I separated this into dy/e^(-2y) = (x-4)dx

I then integrated it to get e^(2y)/2 = x^2/2 - 4x

I then tried to isolate y and got y= 1/2ln(x^2-8x+C)

Plugging in the values, I get ln(4)=1/2ln(-16+C)

What do I do with that 1/2 out in front? If it wasn't a part of the equation, C would equal 20. That is, if I'm correct so far.

Last edited: Jan 30, 2012
2. Jan 30, 2012

### Simon Bridge

Put the everything inside the ln - it will make you feel happier.

$$e^{2y} = x^2-8x+c \Rightarrow y=\ln \sqrt{x^2-8x+c}$$... now the initial condition makes sense?

3. Jan 30, 2012

### Dick

You add the +C after you integrate. Not at the very end. So you should have gotten e^(2y)/2 = x^2/2 - 4x + C. You can actually solve for C at that step as well. Then isolate y.

4. Jan 30, 2012

### TrueStar

Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.

5. Jan 30, 2012

### Simon Bridge

$$a\ln(x)=ln(x^a)$$
But solve for c while y is still in the exponential like Dick says also works.

6. Jan 30, 2012

### Dick

I'm not sure what's throwing you off. C isn't 20. It's something else. Solve for it!

7. Jan 30, 2012

### TrueStar

I forgot the rule that Simon posted. It makes complete sense now. I will look at it the other way as well. Thank you both!