Question

Fun first order differential equation question, can anyone help please? :)

Answer 1

you have the solution already right ?

Answer 2

you just need help figuring out the long term solution ?

Answer 3

Yep, i don't know how to get to the answer underneath - I know how to solve the separable equation but can't get it into the correct form to integrate the left hand side.

Answer 4

Can you show us exactly what's giving you trouble? I'm not sure I understand.

Answer 5

Sure, i've got this far:

\[1/K \int\limits_{?}\frac{ 1 }{ (a-x)(b-x) }{}dx = \int\limits dt \]

Answer 6

familiar with partial fractions ?

Answer 7

and notice that : (a-x)(b-x) = (x-a)(x-b)
doesnt really matter... but just this form looks bit pleasant to me :)

Answer 8

Ah ok, and yep i did have a go at the fractions but i must have been doing something wrong

Answer 9

\[ \int \frac{ 1 }{ (x-a)(x-b) }{}dx = K\int\limits dt \]

Answer 10

\[ \frac{ 1 }{ (x-a)(x-b) } = \frac{A}{x-a} + \frac{B}{x-b} \]

Answer 11

whats ur \(A\) and \(B\) ?

Answer 12

I've got 1= xA-bA+xB-aB ...

Answer 13

heard of coverup method ?

Answer 14

Never heard of it!

Answer 15

nvm then :)
lets just compare the coefficients

Answer 16

\( 1= xA-bA+xB-aB \)

plugin x = 0,
you get : \(1 = -bA -aB~~~~~~\color{red}{*}\)

compare x coefficients ,
you get : \(0 = A + B~~~~~~\color{red}{*}\)

Answer 17

solve them for \(A\) and \(B\)

Answer 18

Ah awesome! Yes i remember plugging numbers in but haven't used that method in a long time. Thanks a lot :)

Answer 19

np :) see if u can manage to get to the solution... interpreting the solution as \(t \to \infty\) can be tricky....

Answer 20

If I may chime in, another way of thinking about it which is just as valid is to say that for:

1=xA-bA+xB-aB we can say that the parts that are multiplied by x on the left must be also multiplied by x on the right and the parts that aren't multiplied by x are equal to those on the other side. Since there are no x's on the left, the x terms will be zero. Think of it by replacing what we're talking with of adding apples and oranges.

1*apples = 6*oranges + x*apples -3y*oranges-4*apples

So we just collect all the stuff. Notice we also have zero oranges on the left, not mysterious though!

0*oranges = 6*oranges - 3y*oranges
1*apples = x*apples-4*apples

So we can just divide out the apples and oranges to get

0=6-3y
1=x-4

We are doing something identical to that here with the x's and non x-terms and I hope you see what I'm talking about!

Answer 21

That makes sense, thank you :)

I can't see where the b/a comes from in the solution?\[\ln((x-b)/(x-a)) = K (t+C)\]
How do we know the limits are from b to a?

Answer 22

its an IVP

Answer 23

you're given \(x(0) = 0 \),
plugin t = 0 , x = 0 in the solution and find out \(C\)

Answer 24

btw, what happened to the factor 1/(b-a) ?

Answer 25

Ah ok so that gives us the b/a ... I'm not sure where 1/b-a went!!

Answer 26

lol u better get sure of it :)

Answer 27

\[\int \frac{ 1 }{ (x-a)(x-b) }dx = K\int\limits dt \]

\[\int \frac{1}{(a-b)(x-a)} - \frac{1}{(a-b)(x-b)} dx = Kt + C\]

\[ \frac{1}{a-b} \ln \left|\frac{x-a}{x-b}\right| = Kt + C \]

Answer 28

Ahhhhh I get it now, that's looking a lot better. I was trying to use a definite integral to get b-a. Now raising to e and plugging in x(0) will give the value of C.
I still dont get how we got that factor of 1/a-b though?

Answer 29

you're okay wid second step above ?

Answer 30

Nope thats the step i dont get, sorry about this!

Answer 31

cool :) thats the step where we changed into partial fractions

Answer 32

what did u get for \(A\) and \(B\) when u had solved them earlier ?

Answer 33

^^

Answer 34

Ok so A=-1 B=1 ....The second equation then gives 1=b-a...

Answer 35

Oh!!! b-a = 1/(x-a) - 1/(x-b)

Answer 36

we have two equations to solve for \(A\) and \(B\) :
\(1 = -bA-aB\)
\(0 = A+B\)

From the second equation, \(B = -A\)
plug this in first equation :
\(1 = -bA + aA \implies A = \dfrac{1}{a-b}\)
so, \(B = \dfrac{1}{b-a}\)

Answer 37

Oh i'm being thick today!! Thank you so much for your help, I get it now :D

Answer 38

\[\frac{ 1 }{ (x-a)(x-b) } = \frac{A}{x-a} + \frac{B}{x-b} \]

\[\frac{ 1 }{ (x-a)(x-b) } = \frac{1}{(a-b)(x-a)} + \frac{1}{(b-a)(x-b)} \]


which is same as :

\[\frac{ 1 }{ (x-a)(x-b) } = \frac{1}{(a-b)(x-a)} - \frac{1}{(a-b)(x-b)} \]

Answer 39

:)

Answer 40

its always good to make sure things

Answer 41

\[\int \frac{ 1 }{ (x-a)(x-b) }dx = K\int\limits dt \]

\[\int \frac{1}{(a-b)(x-a)} - \frac{1}{(a-b)(x-b)} dx = Kt + C\]

\[ \frac{1}{a-b} \ln \left|\frac{x-a}{x-b}\right| = Kt + C ~~~~\color{red}{*}\]


plugin x = 0, y = 0 and solve \(C\)