Hello

I've this equationusing 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 !!!!

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- Feb 12th 2011, 04:58 AMLiverpoolSolve 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 !!!! - Feb 12th 2011, 05:29 AMAckbeet
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

$\displaystyle u'=e^{u+x+e^{x}}=e^{u}e^{x+e^{x}}.$ - Feb 12th 2011, 05:54 AMLiverpool
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, 06:20 AMGeneral
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)

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

Multiplying (2) by $\displaystyle e^{-y}$ and substituting $\displaystyle t=e^{-y}$ will give you a linear equation. - Feb 12th 2011, 06:36 AMAckbeet
- Feb 15th 2011, 08:53 PMLiverpool
Thanks all.