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Thread: DV equation.

  1. #1
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    DV equation.

    Hello,

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

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

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

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

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

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

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


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

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

    $\displaystyle
    \int \frac{1}{y}dy=\int\frac{x+1}{x}dx
    $

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

    Greets.
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  4. #4
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    It seems that, $\displaystyle x\geq 0$
    $\displaystyle y=x^2+Cxe^x$ 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,
    $\displaystyle y'-(1+1/x)y=x-x^2$
    Begin by solving,
    $\displaystyle \frac{dy}{dx}-(1+1/x)y=0$
    Dis is seperable which gives,
    $\displaystyle \frac{1}{y}dy=(1+1/x)dx$
    Integrating,
    $\displaystyle \ln y=x+\ln x+C$
    Thus,
    $\displaystyle y=\exp (x+\ln x+C)=Cxe^x$

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