## Idealist10 one year ago Find the general solution of (x-2)(x-1)y'-(4x-3)y=(x-2)^3.

1. Idealist10

@myininaya

2. Idealist10

@beccaboo333

3. myininaya

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

4. Idealist10

5. myininaya

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

6. Idealist10

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

7. myininaya

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

8. Idealist10

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

9. myininaya

partial fractions

10. Idealist10

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

11. myininaya

no

12. 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

13. Idealist10

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

14. myininaya

... Have you ever done partial fractions before?

15. Idealist10

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

16. 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

17. 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

18. 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

19. myininaya

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

20. myininaya

Anyways back to what you were saying...

21. myininaya

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

22. myininaya

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

23. Idealist10

Yes.

24. myininaya

do you see how that could be helpful here?

25. Idealist10

Yes, wait a second.

26. Idealist10

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

27. myininaya

Very good!

28. Idealist10

Let me work it out then.

29. Idealist10

Y

30. Idealist10

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

31. myininaya

Probably mostly because of the partial fractions part?

32. Idealist10

Yes!

33. 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.

34. myininaya

just practice practice it will sink in eventually