Find the general solution of (x-2)(x-1)y'-(4x-3)y=(x-2)^3.

- Idealist10

Find the general solution of (x-2)(x-1)y'-(4x-3)y=(x-2)^3.

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- Idealist10

@myininaya

- Idealist10

@beccaboo333

- myininaya

so i assumed you tried to put in the form y'+p*y=q form

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## More answers

- Idealist10

Please help me. How do I start? Divide (x-2)?

- myininaya

divide by (x-2)(x-1)

- Idealist10

Okay, I'll try and tell me what's going on. Please don't leave.

- myininaya

you may have to some partial fractions just to let you know

- Idealist10

But how do I integrate (4x-3)/((x-2)(x-1))?

- myininaya

partial fractions

- Idealist10

So do I multiply (x-2)(x-1)? It's x^2-3x+2.

- myininaya

no

- myininaya

\[\frac{3-4x}{(x-2)(x-1) }=\frac{A}{x-2}+\frac{B}{x-1}\]
before you integrate you write what is on the left hand side in the right hand sides's form
you must find A and B to complete that transaction

- Idealist10

So 4x/(x-2)-3/(x-1)? And 4x(x-1)-3(x-2)? But equal to what?

- myininaya

...
Have you ever done partial fractions before?

- Idealist10

So isn't it 5ln(x-2)-ln(x-1)?

- myininaya

I will show an example:
\[\frac{1}{(x-5)(x-3)}\]
pretend we wanted to write this as a sum of proper fractions
we can attempt to do this by trying to write
that fraction in this form A/(x-5)+B/(x-3)
To find what A and B are we will need to combine those fractions in put in to form that is 1/[(x-5)(x-3)] form.
so let's do that
\[\frac{A}{x-5}+\frac{B}{x-3}=\frac{A(x-3)+B(x-5)}{(x-5)(x-3)}=\frac{(A+B)x-3A-5B}{(x-5)(x-3)}\]
but the other side is equal to 1/[(x-5)(x-3)]
so that means the other can have no x's so A+B would have to be 0
and the constant term will have to be equal to 1 therefore -3A-5B=1
so we have a system of linear equations to solve:
A+B=0
-3A-5B=1
-----------
To solve I will try to setup for elimination
Multiply first equation by 3
3A+3B=0
-3A-5B=1
-----------now add
-2B=1
B=-1/2
If B=-1/2 and A+B=0, then A=1/2
So you know that
\[\frac{1}{(x-5)(x-3)} \text{ can be written as } \frac{1}{2} \frac{1}{x-5}-\frac{1}{2}\frac{1}{x-3}\]
and we know how to intgrate both of those terms

- myininaya

@Idealist10 for the integrating factor there was a negative sign in front of y term
so take what you have and multiply it by -1
but don't forget the integrate factor has that base e thing
so for integrate factor we should have
\[v=e^{-5\ln|x-2|+\ln|x-1|}\]
This can be a lot prettier

- myininaya

Now that was just one example above from partial fractions
There are other things to consider when writing a fraction composed of polynomials as a sum of fractions

- myininaya

What I'm saying is the section on partial fractions can not be explained by one example

- myininaya

Anyways back to what you were saying...

- myininaya

to write your integrating factor a lot prettier
you will need to recall some law of exponents and also some log properties

- myininaya

you do recall that x^(a+b) can be written as (x^a)(x^b)
?

- Idealist10

Yes.

- myininaya

do you see how that could be helpful here?

- Idealist10

Yes, wait a second.

- Idealist10

So I got v=(x-1)/(x-2)^5, is this right?

- myininaya

Very good!

- Idealist10

Let me work it out then.

- Idealist10

Y

- Idealist10

Yes, I got it! Thank you so much! Wow, this is a tough problem to solve!

- myininaya

Probably mostly because of the partial fractions part?

- Idealist10

Yes!

- myininaya

I guess you are in differential equations and try to recall all the things you learned in calculus 2.
I do agree I had some trouble but with practice it should become easier on what you should do when you see something like that.

- myininaya

just practice practice
it will sink in eventually

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