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Math Help - solve the given differential by variation of parameters..

  1. #1
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    solve the given differential by variation of parameters..

    ok so the problem is

    xy'' - 4y' = x^4

    im getting a different answer from the book and was wondering if someone could check my work.

    Yc= c1 + c2x^5

    W = 5x^4

    W1 = -x^9

    W2 = x^4


    it seems that the book is using f(x) = x^3???

    i thought the f(x) function would be x^4?

    thanks in advance....
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  2. #2
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    If you let \displaystyle Y = y' then the DE becomes

    \displaystyle xY' - 4Y = x^4

    \displaystyle Y' - \frac{4}{x}Y = x^3.

    The integrating factor is \displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{\left(x^{-4}\right)}} = x^{-4}. Multiplying both sides of the DE by the integrating factor gives

    \displaystyle x^{-4}Y' - 4x^{-5}Y = x^{-1}

    \displaystyle \left(x^{-4}Y\right)' = x^{-1}

    \displaystyle x^{-4}Y = \int{x^{-1}\,dx}

    \displaystyle x^{-4}Y = \ln{|x|} + C_1

    \displaystyle Y = x^4\ln{|x|} + C_1x^4

    \displaystyle y' = x^4\ln{|x|} + C_1x^4

    \displaystyle y = \int{x^4\ln{|x|} + C_1x^4\,dx}

    \displaystyle y = \frac{1}{5}x^5\ln{|x|} - \frac{1}{25}x^5 + \frac{C_1}{5}x^5 + C_2.
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  3. #3
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    thanks for that, i got everything except the C1 term.

    my answer was y= (1/5)x^5ln|x|-(1/25)x^5+x^5C1+C2 <-- ill take another look at the problem.
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  4. #4
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    Quote Originally Posted by slapmaxwell1 View Post
    thanks for that, i got everything except the C1 term.

    my answer was y= (1/5)x^5ln|x|-(1/25)x^5+x^5C1+C2 <-- ill take another look at the problem.
    What we've got only differs by a constant, since \displaystyle \frac{C_1}{5} is still a constant...
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