I am sorry, I cannot provide a title as it would require giving away the answer.

In summary, Gingermom wrote x(y)=-ln(-e^y+1+e) which does not satisfy x(1) = 0, whereas what haruspex originally posted does.
  • #1
gingermom
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Homework Statement





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

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

Homework Equations



it does specifically state dx/dy not dy/dx


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)


 
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  • #2
gingermom said:
Find the particular solution of the differential equation that satisfies the initial condition.

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

The Attempt at a Solution



dx/dy= e^x *e^y
This is different from your original PDE. Did you write it incorrectly?
 
  • #3
strangerep said:
This is different from your original PDE. Did you write it incorrectly?
Looks like gingermom meant e^(x+y).

gingermom, does your solution satisfy the conditions set out in the question?
 
  • #4
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
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
gingermom said:
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.
What you originally wrote would normally be interpreted as [tex]dx/dy= e^x+ y[/tex].

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
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
HallsofIvy said:
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.
No, the given condition is x(1) = 0, not x(0) = 1.
gingermom said:
so the particular solution would be
x(y) = -ln(-e^y+e-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:
1=(e^1+c)^-1
c=e-1
 
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  • #9
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
Sorry for my comment - it is what I get for not paying more attention to the details - everyone's help is greatly appreciated
 

1. What is a particular solution in terms of Dx/dy?

A particular solution in terms of Dx/dy is a solution to a differential equation that satisfies all the given initial conditions. It is typically denoted as yp(x).

2. How is a particular solution different from the general solution?

A particular solution is a specific solution to a differential equation, while the general solution is a family of solutions that includes all possible solutions to the equation. The general solution also includes the particular solution.

3. Can a particular solution be found using different methods?

Yes, a particular solution can be found using different methods such as the method of undetermined coefficients, variation of parameters, or the Laplace transform method. The method used will depend on the type of differential equation and the given initial conditions.

4. Is a particular solution unique?

No, a particular solution is not unique. There can be multiple particular solutions to a given differential equation, depending on the method used to find the solution.

5. How is a particular solution used in practical applications?

A particular solution is used in practical applications to model real-world phenomena and make predictions. By finding a particular solution to a differential equation, we can determine the behavior of a system over time and make informed decisions based on the results.

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