Question

Find the particular solution of the differential equation that satisfies the initial condition. dy/dx = e^(x+y); x(1) = 0

So far I have been able to get the answer e^y = e^x + C or y = ln(C+e^x) but I am not sure how to find C since the initial condition is a bit weird to me.

Answer 1

\[\frac{dy}{dx}=e^xe^y \\ e^{-y} dy=e^x dx\]
I'm not sure you integrate the left hand side correctly

Answer 2

oh you have dx/dy
in the pdf
and dy/dx here

Answer 3

\[\frac{dx}{dy}=e^x e^y \\ e^{-x} dx=e^y dy \]

Answer 4

png whatever :p

Answer 5

but anyways once you integrate that
you have x(a)=b
means to replace y with a and x with b so you can find the constant C

Answer 6

Oh okay! Thanks

Answer 7

So I integrated to get e^y = -e^-x + c simplified to y = ln(-e^-x + c) and C = e+1. Is this correct?

Answer 8

And the solution would be Y = ln(-e^-x + e + 1)

Answer 9

\[-e^{-x}+C=e^y \\ x(1)=0 \text{ \implies we have } \\ -e^{-0}+C=e^{1} \\ -1+C=e \\ C=e+1 \\ \text{ so the solution is } -e^{-x}+e+1=e^y \\ \text{ or you can solve for } y \\ y=\ln (-e^{-x}+e+1)\]
sounds good to me :)