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Thread: 2nd Order ODE

  1. #1
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    2nd Order ODE

    Can anyone take me through solving this 2nd Order ODE?


    $\displaystyle y'' - 4y' + 3y = e^3^x$




    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: 2nd Order ODE

    Quote Originally Posted by remember View Post
    $\displaystyle y'' - 4y' + 3y = e^3^x$
    Just routine knowing the corresponding theory. Have you studied it?
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: 2nd Order ODE

    First calculate the homogenous DE:
    $\displaystyle y''-4y'+3y=0$ and determine the general solution, afterwards you've to find a particular solution.

    Can you do that?
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  4. #4
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    Re: 2nd Order ODE

    I understand that i must first find a complementary function, which i worked out as:

    $\displaystyle y(x)=Ae^x + Be^3^x$ - where A and B are constants.


    My problem is with the next step. I've previously done examples where the homogeneous expression is equal to a quadratic expression... what do i do in the case of an exponential?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: 2nd Order ODE

    That looks good for the general solution, now the particular solution will look like: $\displaystyle a\cdot x \cdot e^{3x}$ (because of $\displaystyle e^{3x}$), you have to find $\displaystyle a$. Calculate the first derivative and second derivative of$\displaystyle a\cdot x \cdot e^{3x}$ and substitute in the DE to find $\displaystyle a$.
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  6. #6
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    Re: 2nd Order ODE

    I have worked out that:

    $\displaystyle y = a e^3^x x$
    $\displaystyle y' = e^3^x (3x + 1)$
    $\displaystyle y'' = 3e^3^x (3x + 2)$


    When you say substitute into the DE, do you mean like this:

    $\displaystyle 3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x$?

    Then expand the brackets and work out a?

    Is my final solution in the form:

    $\displaystyle y(x) = a x e^3^x + Ae^x + Be^3^x $?
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  7. #7
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    Re: 2nd Order ODE

    Quote Originally Posted by remember View Post
    I have worked out that:

    $\displaystyle y = a e^3^x x$
    $\displaystyle y' = e^3^x (3x + 1)$
    $\displaystyle y'' = 3e^3^x (3x + 2)$


    When you say substitute into the DE, do you mean like this:

    $\displaystyle 3e^3^x (3x + 2) - 4(e^3^x (3x + 1) + 3(a e^3^x x) = e^3^x$?

    Then expand the brackets and work out a?

    Is my final solution in the form:

    $\displaystyle y(x) = a x e^3^x + Ae^x + Be^3^x $?
    Yes. Review the theory (method of undetermined coefficients) in your class notes or textbook.
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: 2nd Order ODE

    If you calculate $\displaystyle a$:
    $\displaystyle y=a\cdot x\cdot e^{3x} $
    $\displaystyle y'=a\left(e^{3x}+3x\cdot e^{3x}\right)=a\cdot e^{3x}+3ax \cdot e^{3x}$
    $\displaystyle y''=3a\cdot e^{3x}+3a \left(e^{3x}+3x\cdot e^{3x}\right)$
    $\displaystyle =6a \cdot e^{3x}+9ax\cdot e^{3x}$

    If you substitute this in the equation you get:
    $\displaystyle 6a\cdot e^{3x}+9ax\cdot e^{3x}-4\left(a\cdot e^{3x}+3ax\cdot e^{3x}\right)+3\left(ax\cdot e^{3x}\right)=e^{3x}$
    $\displaystyle \Leftrightarrow 2a\cdot e^{3x}=e^{3x}$
    $\displaystyle \Leftrightarrow a=\frac{1}{2}$

    So the solution of the ODE:
    $\displaystyle y=A\cdot e^{x}+B\cdot e^{3x}+\frac{1}{2}\cdot x\cdot e^{3x}$
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