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Math Help - another DE problem (well more of a parts prob)

  1. #1
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    another DE problem (well more of a parts prob)

    Me agian, catch up on my DE HW...(150ish DE's in 4days ) im doing a linear prob.

    Question: x^2(dy/dx) + x(x+2)y = e^x

    I have found roe (p) and now im trying to intergrate the right side....i come up with... Dx[2x(e^x)y] = 2 (intergal sign)(x^-1)(e^2x)

    ive tryed to use intergrating by parts....but it helps not as i get down to (above intregal) (intregal sign) (-x^(-2))x((1/2x)e^2x) and im back where i started..

    The answer i need to get it to is....y= 1/2(x^-2)e^x+cx^-2(e^-x)


    I'd appreciate any help...also any hints to what i did wrong
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  2. #2
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    x^2 y' + x(x + 2)y = e^x \implies y' + x^{ - 1} (x + 2)y = x^{ - 2} e^x

    Now find the integrating factor.
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  3. #3
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    i got to there....my intergrating factor came out to be....2x(e^x)
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  4. #4
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    I think the integrating factor should be e^xx^2

    Show your process to find it.
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  5. #5
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    the biggest problem i get is when i go to intergrate the right side and use intergration by parts...i get u = x^-1 du= -x^-2 dv= e^2x V = 1/2 e^2x

    but that brings me back to x^-1(1/2 e^2x) (intragal sign) (1/2e^2x)(x^-1)

    which just gives me another parts intregal....almost the same one minus the constant multiplyer which wont really help anything

    not really sure if i just doing my math wrong or which
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  6. #6
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    (x+2)/x -> 1+2/x

    e^(intregal) 1+2/x -> x+ 2ln lxl -> e and ln cancel leaving 2xe^x
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  7. #7
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    \exp \int {\frac{{x + 2}}<br />
{x}~dx} = e^{x + 2\ln \left| x \right|} = e^x \cdot e^{\ln x^2 } = e^x x^2
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  8. #8
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    oh duh.....sorry, i havent used log, ln and "e" in ages....forgot all the rules. know a good place to brezze back up on em?
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  9. #9
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    It's quite dangerous solving differential equations if you don't consider the basic properties.

    I've got to go now, but I hope someone could give you a 'refresh'

    Greetings
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