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 !!!!
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.