# Homework Help: Dx/dy particular solution

1. May 15, 2014

### gingermom

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

Find the particular solution of the differential equation that satisfies the initial condition.

dx/dy = e^x+y x(1) = 0

2. Relevant equations

it does specifically state dx/dy not dy/dx

3. The attempt at a solution

dx/dy= e^x *e^y
1/e^x dx=e^y dy

integrate both sides
-e^-x =e^y +C
e^-x=-e^y+C
-x=ln(-e^y +c)
x=-ln (-e^y+c) but I can't take a ln of -e so how do I find c?

0=-ln(-e^1+c)
0=ln(C-e)
1=c-e Can I do this? Since 0 raised to any power is still zero, I am not sure about whether I can get rid of the Ln this way.
then c= 1+C
If so the particular solution would be
x(y)=-ln(-e^y+1+e)

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

2. Relevant equations

3. The attempt at a solution

2. May 15, 2014

### strangerep

This is different from your original PDE. Did you write it incorrectly?

3. May 15, 2014

### haruspex

Looks like gingermom meant e^(x+y).

gingermom, does your solution satisfy the conditions set out in the question?

4. May 15, 2014

### gingermom

Yes it starts with e^(x+y) but is that not equal to e^x *e^y? that is what I used to separate the variables. If you mean does x(1) =0 fit the equation then yes. If you mean is the derivative of that equal to e^(x+y) - not meaning to be flip, but if I was 100 percent sure I wouldn't have asked the question.

5. May 15, 2014

### gingermom

Did you mean that I found the specific solution in terms of x(y) and the way the question is worded I need to go back and solve that for y(x)? I am not clear on what you are asking when you say does my solution satisfy the question?

6. May 15, 2014

### HallsofIvy

What you originally wrote would normally be interpreted as $$dx/dy= e^x+ y$$.

But haruspex's question was related to "x(0)= 1".

With "x(y)=-ln(-e^y+1+e)" x(0)= -ln(-1+ 1 + e)= -ln(e)= -1, not 1.

7. May 15, 2014

### gingermom

Oh I see what you mean - sorry for the wrong input. I multiplied both sides by -1 to get
0=ln(e^1 +c)
0=ln1
ln1=ln (e^1 +c)
1=e^1 +c
c= 1-e^1 so I should not have multiplied by -1 because both sides are not in ln duh!
so it would be
0=- ln(e^1 +c)
0=ln1
ln1=-1 ln (e^1 +c) -r can I multiple
1=(e^1+c)^-1
c=e-1

so the particular solution would be
x(y) = -ln(-e^y+e-1)

8. May 15, 2014

### haruspex

No, the given condition is x(1) = 0, not x(0) = 1.
But that does not satisfy x(1) = 0, whereas what you originally posted does!
The point of my question in post #3 is that your solution in the OP does satisfy all the given conditions and is therefore the right answer.
Your mistake in post #7 is here:

9. May 15, 2014

### gingermom

Well I guess I will know soon - since I turned in the corrected answer rather than my original post - guess that is what I guess for asking.

10. May 15, 2014

### gingermom

Sorry for my comment - it is what I get for not paying more attention to the details - everyone's help is greatly appreciated