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Idealist10

  • 4 months ago

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

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  1. Idealist10
    • 4 months ago
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    @myininaya

  2. Idealist10
    • 4 months ago
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    @beccaboo333

  3. myininaya
    • 4 months ago
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    so i assumed you tried to put in the form y'+p*y=q form

  4. Idealist10
    • 4 months ago
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    Please help me. How do I start? Divide (x-2)?

  5. myininaya
    • 4 months ago
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    divide by (x-2)(x-1)

  6. Idealist10
    • 4 months ago
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    Okay, I'll try and tell me what's going on. Please don't leave.

  7. myininaya
    • 4 months ago
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    you may have to some partial fractions just to let you know

  8. Idealist10
    • 4 months ago
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    But how do I integrate (4x-3)/((x-2)(x-1))?

  9. myininaya
    • 4 months ago
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    partial fractions

  10. Idealist10
    • 4 months ago
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    So do I multiply (x-2)(x-1)? It's x^2-3x+2.

  11. myininaya
    • 4 months ago
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    no

  12. myininaya
    • 4 months ago
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    \[\frac{3-4x}{(x-2)(x-1) }=\frac{A}{x-2}+\frac{B}{x-1}\] 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
    • 4 months ago
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    So 4x/(x-2)-3/(x-1)? And 4x(x-1)-3(x-2)? But equal to what?

  14. myininaya
    • 4 months ago
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    ... Have you ever done partial fractions before?

  15. Idealist10
    • 4 months ago
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    So isn't it 5ln(x-2)-ln(x-1)?

  16. myininaya
    • 4 months ago
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    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
    • 4 months ago
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    @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
    • 4 months ago
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    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
    • 4 months ago
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    What I'm saying is the section on partial fractions can not be explained by one example

  20. myininaya
    • 4 months ago
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    Anyways back to what you were saying...

  21. myininaya
    • 4 months ago
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    to write your integrating factor a lot prettier you will need to recall some law of exponents and also some log properties

  22. myininaya
    • 4 months ago
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    you do recall that x^(a+b) can be written as (x^a)(x^b) ?

  23. Idealist10
    • 4 months ago
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    Yes.

  24. myininaya
    • 4 months ago
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    do you see how that could be helpful here?

  25. Idealist10
    • 4 months ago
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    Yes, wait a second.

  26. Idealist10
    • 4 months ago
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    So I got v=(x-1)/(x-2)^5, is this right?

  27. myininaya
    • 4 months ago
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    Very good!

  28. Idealist10
    • 4 months ago
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    Let me work it out then.

  29. Idealist10
    • 4 months ago
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    Y

  30. Idealist10
    • 4 months ago
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    Yes, I got it! Thank you so much! Wow, this is a tough problem to solve!

  31. myininaya
    • 4 months ago
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    Probably mostly because of the partial fractions part?

  32. Idealist10
    • 4 months ago
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    Yes!

  33. myininaya
    • 4 months ago
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    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
    • 4 months ago
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    just practice practice it will sink in eventually

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