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Math Help - 2nd order DE's with constant coefficients.

  1. #1
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    Post 2nd order DE's with constant coefficients.

    1- How can I solve this equation? y''+4y'=0

    2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
    e^(3x) and xe^(3x)

    Thanks for your help.
    Last edited by mr fantastic; January 9th 2011 at 01:24 PM. Reason: Re-titled.
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    Quote Originally Posted by hejmh View Post
    1- How can I solve this equation? y''+4y'=0

    2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
    e^(3x) and xe^(3x)

    Thanks for your help.
    y''+4y'=0\Rightarrow m^2+4m=0

    Do you know how to use this form?
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  4. #4
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    For part 2, you're told \displaystyle y = e^{3x} is a solution.

    So \displaystyle y' = 3e^{3x} and \displaystyle y'' = 9e^{3x}.

    Substitute into the DE to get \displaystyle 9e^{3x} + 3a\,e^{3x} + b\,e^{3x} = 0.


    Now do the same with the other solution \displaystyle y = x\,e^{3x} to get another equation that involves \displaystyle a and \displaystyle b. Then you can solve the two equations simultaneously for \displaystyle a and \displaystyle b.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    y''+4y'=0\Rightarrow m^2+4m=0

    Do you know how to use this form?
    From this point, you factor and solve for m.

    m(m+4)=0\Rightarrow m=0,-4

    y=C_1e^{m_1x}+C_2e^{m_2x}

    y=C_1e^{0*x}+C_2e^{-4x}=C_1+C_2e^{-4x}
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    Yes. thanks
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  7. #7
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    Quote Originally Posted by hejmh View Post
    1- How can I solve this equation? y''+4y'=0
    Here's another way to do this. Since there is no "y" (undifferentiated) let u= y' so the equation becomes u'+ 4u= 0. That is the same as du/dx= -4u which is a separable, first order differential equation. You can write it as du/u= -4 dx and integrate both sides: ln(u)= -4x+ c which leads to u(x)= y'(x)= e^{-4x+ c}= Ce^{-4x} where C= e^{c}. Now find y by integrating y'= Ce^{-4x}: y(x)=  \frac{C}{-4}e^{4x}+ B which we can also write as y(x)= Ae^{-4x}+ B where now A= \frac{C}{-4}= \frac{e^c}{-4}.
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