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Linear diffy q

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation
    x(dy/dx) -4y = x^4e^x

    2. Relevant equations


    3. The attempt at a solution


    So this is what I did.

    dy/dx -4y/x = x^3e^x

    Then I did the integral of P(x) which I said was 4/x so the integral is lnx^4 then I(x) = e^lnx^4 = x^4. I then multiplied through by this and got


    x^4(dy/dx) - 4x^3(y) = x^7e^x I said my product rule for my left side was (x^4(y))' so I ended up with x^4(y) = ∫ x^7(e^x) I just used a reduction formula for the integral. But is my procss OK? THanks
    my equation b
     
  2. jcsd
  3. Oct 30, 2013 #2

    pasmith

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    In this case [itex]P(x) = -4/x[/itex], so you've lost a minus sign in an exponent.

    Try again with the correct [itex]P(x)[/itex].
     
  4. Oct 30, 2013 #3
    Hey dude, So I did but it doesn't seem right I made P(x) as you say so my integrating factor become I(x) = 1/x^4

    I multiply though and my expression becomes

    (1/x^4)(dy/dx) - 4y/x^5 = e^x/x OK so the left side is then....

    (y/x^4)' = e^x /x
    But if you try and integrate the right side you have to do the err function. That can't be right because my book isn't that advanced. So Did I do something wrong?
     
  5. Oct 30, 2013 #4

    Dick

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    No, you did everything right. And yes, you don't get an elementary integral. It's not err it's Ei, but same problem.
     
  6. Oct 30, 2013 #5
    OK. I kept looking over and it and I could not see where my math was wrong you know?
    Hey I'm going to post another thread can you please help me ? It is solving a differential equation.
     
  7. Oct 30, 2013 #6

    Dick

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    Sure, if I'm still around. Somebody else could probably handle it too.
     
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