# Solve the following ODE .. #9

• Feb 12th 2011, 05:58 AM
Liverpool
Solve the following ODE .. #9
Hello

I've this equation using inspection :

$\displaystyle \dfrac{dy}{dx} = 1 + e^{y+e^x}$

ok
I couldn't solve it using any method !!
how in earth can i solve this one using inspection !!!!
• Feb 12th 2011, 06:29 AM
Ackbeet
Well, I don't know about inspection, but the substitution $u=y-x$ renders the equation separable:

$u'=y'-1,$ and hence the DE becomes

$u'=e^{u+x+e^{x}}=e^{u}e^{x+e^{x}}.$
• Feb 12th 2011, 06:54 AM
Liverpool
Thanks !!
it is really hard
thanks you man
But am really intrested for the "inspection solution" for it since it is in the inspection section's exercises
• Feb 12th 2011, 07:20 AM
General
Hello,

Here is what do you want:

We have $\dfrac{dy}{dx}=1+e^y \cdot e^{e^x}$

$\implies dy=dx+e^y \cdot e^{e^x} \, dx$

$\implies e^x \, dy = e^x dx + e^x e^y e^{e^x} \, dx$

$\implies \dfrac{e^x}{e^y} \, dy = \dfrac{e^x}{e^y} \, dx + e^x e^{e^x} \, dx$

$\implies e^{x-y} \, dy - e^{x-y} dx = e^x e^{e^x} \, dx$

$\implies e^{x-y} \, ( dy - dx ) = e^x e^{e^x} \, dx$

$\implies e^{x-y} \, d(y-x) = e^x e^{e^x} \, dx$

$\implies e^{-(y-x)} \, d(y-x) = e^x e^{e^x} \, dx$ ... (1)

Again, equation (1) is equivalent to the last equation in Ackbeet's reply.

Another way ,if you are intersted , :

We have $\dfrac{dy}{dx}=1+e^y \cdot e^{e^x}$ ... (2)

Multiplying (2) by $e^{-y}$ and substituting $t=e^{-y}$ will give you a linear equation.
• Feb 12th 2011, 07:36 AM
Ackbeet
Quote:

Originally Posted by Liverpool
Thanks !!
it is really hard
thanks you man
But am really intrested for the "inspection solution" for it since it is in the inspection section's exercises

You're quite welcome. I'm not smart enough to "see" a "by inspection solution" for this DE. In looking at the results I get using my method, the solution is much too complicated for me to have "seen it".
• Feb 15th 2011, 09:53 PM
Liverpool
Thanks all.