1. ## Stuck at e^(e^2x)..

Hi, I am a computer engineering student and Maths is one of my subject for this university year.

I was solving some paper and I cant figure out how to solve the following step. (Its based on Linear Differential Equations.)

This is the step I am stuck at,
$\displaystyle e^{-2x} e^{e^{2x}}$

The answer of this step is
$\displaystyle e^{4x}$

How?

2. Originally Posted by enli
Hi, I am a computer engineering student and Maths is one of my subject for this university year.

I was solving some paper and I cant figure out how to solve the following step. (Its based on Linear Differential Equations.)

This is the step I am stuck at,
$\displaystyle e^{-2x} e^{e^{2x}}$

The answer of this step is
$\displaystyle e^{4x}$

How?
You don't even say what you're trying to do with this expression.

3. Originally Posted by mr fantastic
You don't even say what you're trying to do with this expression.

Hi, sorry for the misunderstanding.

I thought the whole problem is not important, so i didnt post it.

In the problem, it is shown that those two expressions are equal. I wish to know how it is done.

To reframe my question, how do you show,

$\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}$

4. Originally Posted by enli
Hi, sorry for the misunderstanding.

I thought the whole problem is not important, so i didnt post it.

In the problem, it is shown that those two expressions are equal. I wish to know how it is done.

To reframe my question, how do you show,

$\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}$
Clearly the expressions are not equal in general. If you have obtained this expression while attempting to solve your original problem, then you have made a mistake. Which is why I asked you to post the original question.

5. The original problem is to solve the following (Linear differential equations)
$\displaystyle (D^2+6D+8)y = e^{e^{2x}}$

The book has solved all steps correctly prior to the following step, where I cant follow how it is done
$\displaystyle \frac{1}{D+4} [e^{-2x}\frac{1}{2}e^{e^{2x}}]$

And the next step is,
$\displaystyle =\frac {1}{2}e^{-4x}\int{ e^{4x}e^{-2x}e^{e^{2x}} dx}$

6. Originally Posted by enli
The original problem is to solve the following (Linear differential equations)
$\displaystyle (D^2+6D+8)y = e^{e^{2x}}$

The book has solved all steps correctly prior to the following step, where I cant follow how it is done
$\displaystyle \frac{1}{D+4} [e^{-2x}\frac{1}{2}e^{e^{2x}}]$

And the next step is,
$\displaystyle =\frac {1}{2}e^{-4x}\int{ e^{4x}e^{-2x}e^{e^{2x}} dx}$
To solve $\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx$ make the substitution $\displaystyle u = e^{2x}$.

7. Originally Posted by mr fantastic
To solve
$\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx$ make the substitution $\displaystyle u = e^{2x}$.
Excuse me for not being clear enough about the problem,

I wish to know following two are equal
$\displaystyle \frac {1}{D+4} [e^{-2x} \frac {1}{2}e^{e^{2x}}] = \frac {1}{2}e^{-4x}\int { e^{4x}e^{-2x}e^{e^{2x}} dx}$

The formula is,
$\displaystyle \frac {1}{D-m}f(x) = e^{mx}\int {e^{-mx} f(x)dx}$

8. Originally Posted by mr fantastic
To solve
$\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx$ make the substitution $\displaystyle u = e^{2x}$.
Yeah right,

The formula is,
$\displaystyle \frac {1}{D+m}f(x) = e^{-mx}\int {e^{mx} f(x)dx}$

Now if we follow the formula, that means
$\displaystyle m=4$
$\displaystyle f(x)=e^{-2x}e^{e^{2x}}$

And hence now it makes sense why I was trying to compare, which was of course wrong
$\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}$