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Thread: ODE

  1. #1
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    ODE

    I re posted because i stuffed up the latex code.


    I am stuck on this ODE.

    Boundary conditions are:
    U(0)=U(1)=0

    $\displaystyle x^2$ $\displaystyle *$$\displaystyle \frac{d^2u}{dx^2}$ +$\displaystyle x$ *$\displaystyle \frac{du}{dx}$ -$\displaystyle 4u$ =$\displaystyle 3x$
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  2. #2
    MHF Contributor chisigma's Avatar
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    For the general solution of the 'incomplete' DE...

    $\displaystyle x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x =0$ (1)

    ... try functions like $\displaystyle u(x)= x^{\alpha}$.For the particular solution of the 'complete' DE...

    $\displaystyle x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x = 3\cdot x$ (2)

    ... try a function like $\displaystyle u(x)= \beta\cdot x$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Quote Originally Posted by ulysses123 View Post
    I re posted because i stuffed up the latex code.


    I am stuck on this ODE.

    Boundary conditions are:
    U(0)=U(1)=0

    $\displaystyle x^2$ $\displaystyle *$$\displaystyle \frac{d^2u}{dx^2}$ +$\displaystyle x$ *$\displaystyle \frac{du}{dx}$ -$\displaystyle 4u$ =$\displaystyle 3x$
    This is an "Euler-type" or "equipotential" equation. As chisigma said, you can try a solution of the form $\displaystyle y= x^\alpha$ to get the "characteristic equation".

    It is also true that the substitution t= ln(x) converts an "Euler-type" equation into an equation with constant coefficients having the same characteristic equation.
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