# Math Help - Seperation of Variables

1. ## Seperation of Variables

Hello, I was wondering if you guys could quickly check over my work to make sure I did this question correctly. We're solving differential equations using the seperation of variables:
y(0)=e
dy/dt = e^(t-y)
ln(dy/dt)=t-y
y+ln(dy)=t+ln(dt)
e^(y+ln(dy))=e^(t+ln(dt))
e^y(dy)=e^t(dt)
(Take integrals of both sides)
e^y=(e^t)+C
ln(e^y)=ln((e^t)+C)
y=ln(e^t+C)
y(0)=e
e=ln(e^0+C)
e=ln(1+C)
e^e=1+C
C=e^e-1
y=ln(e^t+e^e-1)

I'm not sure if this is right, I wasn't sure how to isolate dy/dt so I put both sides to the power of e, I was just wondering if that was the correct thing to do. All help is appreciated!

$\displaystyle \frac{dy}{dt} = e^{t-y}$

$\displaystyle \frac{dy}{dt} = \frac{e^t}{e^y}$

$\displaystyle e^y~dy = e^t~dt$

$\displaystyle \int e^y~dy = \int e^t~dt$

Hello, I was wondering if you guys could quickly check over my work to make sure I did this question correctly. We're solving differential equations using the seperation of variables:

e^y(dy)=e^t(dt)
(Take integrals of both sides)
e^y=(e^t)+C
ln(e^y)=ln((e^t)+C)
y=ln(e^t+C)
y(0)=e
e=ln(e^0+C)
e=ln(1+C)
e^e=1+C
C=e^e-1
y=ln(e^t+e^e-1)

I'm not sure if this is right, I wasn't sure how to isolate dy/dt so I put both sides to the power of e, I was just wondering if that was the correct thing to do. All help is appreciated!
Correct.