Question

Find a solution to the first-order ODE: y'= e^(x-y), y(0)=1

Answer 1

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

Answer 2

Can you do the rest ?

Answer 3

Yes, thank you!! As soon as u typed that, i found it in my notes. There is another one I may need help with. Be on the look out pleeeease!!!

Answer 4

k

Answer 5

Im stuck at \[\int\ e ^{y} dy = \int\ e ^{x} dx\]
I know that \[\int\limits_{?}^{?} e^x =e^x + C\]
Can I do the same for e^y

Answer 6

?

Answer 7

\[\int\limits_{}^{} e^* d * =e^*+C\]

Answer 8

It doesn't matter if * is x or y

Answer 9

\[\int\limits_{}^{}e^y dy=e^y+k\]

Answer 10

Oh, duh, lol. I keep making things more difficult.
So ln(e^y+ k)= ln(e*+c)
y= ln(e*+c)-k ?

Answer 11

\[e^y+k=e^x+C\]
\[e^y=e^x+C-k\]
C-k is still a constant
\[e^y=e^x+c\]
I would leave it like this
But if you are asked to solve for y
Yes take ln( ) of both sides
\[\ln(e^y)=\ln(e^x+c)\]
\[y \ln(e)=\ln(e^x+c)\]
\[y=\ln(e^x+c)\]

Answer 12

Yes, im asked to solve for y, with y(0)=1.
So 1= ln(e^(0)+c)
I know that e^(0)=1, but ln(1)= 0.

Answer 13

\[e^1=e^0+c => e=1+c => e-1=c\]

Answer 14

Thaaaank yoou, oh kind one!!