Question

Solve the IVP:
dy/dt=te^(t-y)
y(0)=5

Answer 1

Looks separable to me.

Answer 2

How so?

Answer 3

e^(t-y) = e^t . e^-y

Answer 4

Oh duh forgot that property :)

Answer 5

so separating them looks like this: e^y dy=te^t dt

Answer 6

then integrate both sides and you get e^y +\[c_{1}\]=te^t-e^t +\[c_{2}\]

Answer 7

Or more simply,
\[ e^y = (t-1)e^x + C \]
thus y = ...

Answer 8

combine c's and get e^y+\[c _{3}\]=te^t-e^t

Answer 9

I mean (t-1)e^t + C

Answer 10

\[ e^y = (t-1)e^t + C \]
thus y = ...

Answer 11

y=tln(t-1)+\[c _{4}\]

Answer 12

No, you can't take the constant outside.

Answer 13

ln(a + b) does not equal ln a + ln b

Answer 14

hence
\[ y = \ln( (t-1)e^t + C) \]

Answer 15

okay so it is y=ln(e^t(t-1)+\[c _{3}\])

Answer 16

and plugging in y(0)=5 gives me \[c _{3}\]=e^5 ???

Answer 17

Indeed

Answer 18

so the final answer would be: y=ln(e^t(t-1)+e^5)

Answer 19

You can verify this by making sure it satisfies the original equation.

Answer 20

opps \[c _{3}\]=e^5 +1

Answer 21

and it works :)

Answer 22

good