Question

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

Answer 1

@myininaya

Answer 2

@beccaboo333

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

partial fractions

Answer 10

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

Answer 11

no

Answer 12

\[\frac{3-4x}{(x-2)(x-1) }=\frac{A}{x-2}+\frac{B}{x-1}\]
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

Answer 13

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

Answer 14

...
Have you ever done partial fractions before?

Answer 15

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

Answer 16

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

Answer 17

@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

Answer 18

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

Answer 19

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

Answer 20

Anyways back to what you were saying...

Answer 21

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

Answer 22

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

Answer 23

Yes.

Answer 24

do you see how that could be helpful here?

Answer 25

Yes, wait a second.

Answer 26

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

Answer 27

Very good!

Answer 28

Let me work it out then.

Answer 29

Y

Answer 30

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

Answer 31

Probably mostly because of the partial fractions part?

Answer 32

Yes!

Answer 33

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.

Answer 34

just practice practice
it will sink in eventually