Question

Solve the initial value problem

Answer 1

\[\frac{ du }{ dt }=tu-t+u-1\] u(0)=2

Answer 2

i multiplied by dt on both sides then i divide by u and subtract a u from both sides?

Answer 3

It's seperable. Factor the right hand side

Answer 4

t(u-1+u/t-1/t)?

Answer 5

No.
tu−t+u−1
t(u-1)+(u-1)
This is the key thing to notice. At times you may want to add and subtract a value to attain this kind of arrangement
(t+1)(u-1)

Answer 6

I finally remember. It's called the AC method of factoring

Answer 7

\[du=(t+1)(u-1)dt\] is that what it will look like?

Answer 8

sorry i just got really confused

Answer 9

Now take all the stuff with u to the left

Answer 10

so it would be like \[\frac{ du }{ u-1 }=(t+1)dt\]

Answer 11

Yup.

Answer 12

then find the integral of both sides right

Answer 13

Yes. You can use bounds if you want or you can get the general solution then solve for the constant of integration based on your initial value

Answer 14

i get the general solution to be\[\ln \left| u-1 \right|=\frac{ t^2 }{ 2 }+t+C\]

Answer 15

then i plug in the IC u(0)=2 (0,2) plugging in 2 for u and ) for t, correct?

Answer 16

0 for t

Answer 17

Yes. You should be find from here.
I'll head on to sleep 2AM ~_~
Goodnight

Answer 18

Goodnight.
Thanks for all the help