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

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

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

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Please help me. How do I start? Divide (x-2)?
divide by (x-2)(x-1)
Okay, I'll try and tell me what's going on. Please don't leave.
you may have to some partial fractions just to let you know
But how do I integrate (4x-3)/((x-2)(x-1))?
partial fractions
So do I multiply (x-2)(x-1)? It's x^2-3x+2.
no
\[\frac{3-4x}{(x-2)(x-1) }=\frac{A}{x-2}+\frac{B}{x-1}\] 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
So 4x/(x-2)-3/(x-1)? And 4x(x-1)-3(x-2)? But equal to what?
... Have you ever done partial fractions before?
So isn't it 5ln(x-2)-ln(x-1)?
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
@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
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
What I'm saying is the section on partial fractions can not be explained by one example
Anyways back to what you were saying...
to write your integrating factor a lot prettier you will need to recall some law of exponents and also some log properties
you do recall that x^(a+b) can be written as (x^a)(x^b) ?
Yes.
do you see how that could be helpful here?
Yes, wait a second.
So I got v=(x-1)/(x-2)^5, is this right?
Very good!
Let me work it out then.
Y
Yes, I got it! Thank you so much! Wow, this is a tough problem to solve!
Probably mostly because of the partial fractions part?
Yes!
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.
just practice practice it will sink in eventually

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