Question

Separable ODE question

Answer 1

I am currently stuck on part(ii) of this question. I've tried using both equations are tired separating them to solve for x, however it's not working out, particularly the integration. I know I am slipping up somewhere

Answer 2

No I was trying to get the x's on side of the equation and the t's on the other.

Answer 3

u were trying to separate it?

Answer 4

Yes

Answer 5

i don't think that's possible

Answer 6

I am trying to find the solution in terms of x, that's what the question is asking me.

Answer 7

helder_edwin - I believe all that is required here is to integrate what is given in part (i)

Answer 8

as it is part (ii) that ironictoaster is stuck on

Answer 9

ok i get it

Answer 10

I tried using the rewritten equation by dividing t^2-1 and multiplying across by dt.

Answer 11

just integrate both sides with respect to t

Answer 12

use the fact that:\[\int\frac{d}{dt}f(t)dt=f(t)\]

Answer 13

i.e.:\[\int\frac{d}{dt}[(t^2-1)x]dt=(t^2-1)x\]

Answer 14

so you just need to integrate the right hand side here

Answer 15

i.e.:\[\int(t-1)\cos(t)dt\]

Answer 16

do you know how to do that?

Answer 17

so
\[ \large \int d[(t^2-1)x]=\int(t-1)\cos t\,dt \]

Answer 18

Ohhh, I did not know I integrate the LHS like that cause there's a t^2-1 there. I divided across before integrating.

Answer 19

@asnaseer Looks like integration by parts.

Answer 20

remember the LHS is a derivative with respect to t.

Answer 21

yes - integration by parts for RHS

Answer 22

sorry what is LHS?

Answer 23

Left hand side

Answer 24

when using integrating factors you get a left hand side (LHS) which is just a derivative with respect to the main variable (t in this case)

Answer 25

that's the advantage of using integrating factors

Answer 26

ohhh
sorry english is not my mother language

Answer 27

you end up with a very simple integral on the LHS

Answer 28

don't forget to include the constant in the integration of the RHS.
then just use the initial values to solve for that constant.

Answer 29

Yeah that's how I did part i. The LHS is always your integrating factor multiplied for a variable and doing the product rule will give the LHS of the previous step.

Answer 30

by a variable*

Answer 31

yes - ok - I believe you know how to solve this now?

Answer 32

Yep sure do, thanks for the help guys!

Answer 33

yw :)