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Math Help - integrating factor or d.e

  1. #1
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    integrating factor or d.e

    when im doing questions that use integrating factors i always face the same problem.

    eg.
    dy/dx + P(x)y = Q(x)
    and the integrating factor, I, would be e^(integral of P)

    however after multiplying the whole equation by I, you would normally get something like

    Iy' + IP(x)y = IQ(x)

    and the next step would be combining the left hand side into (###)' so that you could equate ### to integral of the right hand side. what i want to ask is how do you get from Iy' + IP(x)y to first derivative of ###?

    and example is
    dx/dt + 4x/(100-t) = 3/2

    im not sure if its correct but i got
    P = 4/100-t
    => I = (100-t)^-4 (correct this if im wrong)

    then the equation becomes

    x'(100-t)^-4 + 4x(100-t)^-5 = 3/2(100-t)^-4

    what do i do from here? i always get stuck at this step for all problems using integrating factors.
    Last edited by yen yen; June 6th 2010 at 10:07 PM.
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  2. #2
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    Quote Originally Posted by yen yen View Post
    when im doing questions that use integrating factors i always face the same problem.

    eg.
    dy/dx + P(x)y = Q(x)
    and the integrating factor, I, would be e^(integral of P)

    however after multiplying the whole equation by I, you would normally get something like

    Iy' + IP(x)y = IQ(x)

    and the next step would be combining the left hand side into (###)' so that you could equate ### to integral of the right hand side. what i want to ask is how do you get from Iy' + IP(x)y to first derivative of ###?

    and example is
    dx/dt + 4x/(100-t) = 3/2

    im not sure if its correct but i got
    P = 4/100-t
    => I = (100-t)^4 (correct this if im wrong)

    then the equation becomes

    x'(100-t)^4 + 4x(100-t)^3 = 3/2(100-t)^4

    what do i do from here? i always get stuck at this step for all problems using integrating factors.

    I think you have an error in your integrating factor

    exp(\int Pdt)=exp(\int\frac{4}{100-t}dt)=exp(-4\ln(100-t))=\frac{1}{(100-t)^4}
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    I think you have an error in your integrating factor

    exp(\int Pdt)=exp(\int\frac{4}{100-t}dt)=exp(-4\ln(100-t))=\frac{1}{(100-t)^4}
    yea i corrected it.. but i still dont know what the next step is supposed to be..

    never mind. i figured out how to get it. i only realize now thats what the integrating factor is for.
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  4. #4
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    Quote Originally Posted by yen yen View Post
    yea i corrected it.. but i still dont know what the next step is supposed to be..
    Now multiply the equation by the integrating factor to get

    \frac{1}{(100-t)^4}\frac{dx}{dt}+\frac{4}{(100-t)^5}x=\frac{3}{2}\frac{4}{(100-t)^4}

    Now the left hand side is the derivative of a product and can be written as

    \frac{d}{dt}\left( \frac{1}{(100-t)^4}x\right)=\frac{3}{2}\frac{4}{(100-t)^4}

    Now just integrate both sides with respect to t to solve for x.
    Last edited by TheEmptySet; June 6th 2010 at 10:27 PM. Reason: typo
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