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Math Help - DV equation.

  1. #1
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    DV equation.

    Hello,

    Given is xy^{'}- (x+1)y=x^2-x^3

    Before I start I multiply by \frac{1}{x}

    Then I get y'-\frac{x+1}{x}y=\frac{x^2-x^3}{x}

    First I will find the solution of the homogeneous DV y'-\frac{x+1}{x}y=0

    I can solove this easely \int dy=\int\frac{x+1}{x} and I get y=x+\ln x+k(x)

    I deveritate this and I get y'=1+\frac{1}{x}+k'(x) and now I need to put in but then I miss ???

    Who can help me? Greets.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Bert


    First I will find the solution of the homogeneous DV y'-\frac{x+1}{x}y=0

    I can solove this easely \int dy=\int\frac{x+1}{x} and I get y=x+\ln x+k(x)
    except you have:

    <br />
\int \frac{1}{y}dy=\int\frac{x+1}{x}dx<br />

    RonL
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  3. #3
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    do I need to integrate y'-\frac{x+1}{x}y and not y'-\frac{x+1}{x} ???

    Greets.
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  4. #4
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    It seems that, x\geq 0
    y=x^2+Cxe^x are all solutions to this differential equation.
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  5. #5
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    That one I solved by using a theorem about linear differencial equations. I have another way.

    "The general solution to a non-homogenous differencial equation is the sum of the general solution to the homogenous equation and a specific solution to the non-homogenous equation"

    With that theorem,
    We will solve,
    y'-(1+1/x)y=x-x^2
    Begin by solving,
    \frac{dy}{dx}-(1+1/x)y=0
    Dis is seperable which gives,
    \frac{1}{y}dy=(1+1/x)dx
    Integrating,
    \ln y=x+\ln x+C
    Thus,
    y=\exp (x+\ln x+C)=Cxe^x

    Now find a particular solution to,
    y'-(1+1/x)y=x-x^2
    It is reasonable to search for solutions of the form,
    y=ax^2+bx+c
    Substitute,
    2ax+b-(1+1/x)(ax^2+bx+c)=x-x^2
    Thus,
    2ax+b-ax^2-bx-c-ax-b-c/x=x-x^2
    We immediately see that c=0
    Thus, you are left with,
    ax-ax^2-bx=x-x^2
    We also see that a=1
    From here we see that b=0
    Thus, y=1x^2+0x+0=x^2 is a specific solution.
    Thus, all solutions are,
    y=x^2+Cxe^x
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