# Thread: Solve the following ODE .. #9

1. ## Solve the following ODE .. #9

Hello

I've this equation using inspection :

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

ok
I couldn't solve it using any method !!
how in earth can i solve this one using inspection !!!!

2. Well, I don't know about inspection, but the substitution $\displaystyle u=y-x$ renders the equation separable:

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

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

3. 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

4. Hello,

Here is what do you want:

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

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

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

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

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

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

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

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

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

Another way ,if you are intersted , :

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

Multiplying (2) by $\displaystyle e^{-y}$ and substituting $\displaystyle t=e^{-y}$ will give you a linear equation.

5. 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".

6. Thanks all.