Solve the following ODE .. #9

**Liverpool** · February 12th 2011, 04:58 AM

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 !!!!

**Ackbeet** · February 12th 2011, 05:29 AM

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

$u'=e^{u+x+e^{x}}=e^{u}e^{x+e^{x}}.$

**Liverpool** · February 12th 2011, 05:54 AM

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

**General** · February 12th 2011, 06:20 AM

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)

Your equation is ready for integrating now.
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.

**Ackbeet** · February 12th 2011, 06:36 AM

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".

**Liverpool** · February 15th 2011, 08:53 PM

Thanks all.

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