Question

Solve the differential equation with the initial condition.
dy/dx = e^(-x), y(1) = 3

Answer 1

use separation of variables

Answer 2

I know the integral of e^(-x) is -e^-x

Answer 3

\[\int\limits dy = \int\limits \frac{ 1 }{ e^x } dx\]

Answer 4

plus constant of integration

Answer 5

yeah, so I am at 3 = -e^(-x) + C

Answer 6

you could take a natural log of both sides

Answer 7

sorry, I mean 3 = -e^(-1) + C

Answer 8

Do what @tomo said. Take the natural log and solve for C.
When you solve for C, plug it in.

Answer 9

so ln(3) = ln(-e^(-1)) + C then solve for C?

Answer 10

yes

Answer 11

so I know ln(e^x) is x, but what about ln(-e^(-x))?

Answer 12

ln(-e^(-x)) is not valid because e^(-x)>0
so -e^(-x) <0

Answer 13

Actually ln is not necessary.

ok, so
dy/dx = e^-x
dy = 3^-x*dx
y = integral(1/e^x)dx
y = -e^-x+c
y = c-e^-x

y(1) = 3

ok, so we need to figure out c

3 = c-e^-1

3+(1/e) = c