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Math Help - Need help with solving Non-homogenous 2Nd ODE Eqn.

  1. #1
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    Need help with solving Non-homogenous 2Nd ODE Eqn.

    y''+7y'-8y = e^x y(0)= 8/9 y'(0)=1


    Working will be much appreciated.

    Thanks a million!
    Say Khong.
    God Bless.
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  2. #2
    Member kjchauhan's Avatar
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    y'' + 7y' -8y = e^x

    Auxiliary equation is

    D^2+7D-8=0

    \therefore D=-8, 1

    \therefore C.F.= c_1 e^{-8x} + c_2 e^{x}

    Now P.I. = \frac{1}{D^2+7D-8} e^x =\frac{1}{9}xe^x

    \therefore y(x) = C.F. + P.I.

    \therefore y(x) = c_1 e^{-8x} + c_2 e^{x} + \frac{1}{9} xe^x

    Now,

    y'(x) = -8 c_1 e^{-8x} + c_2 e^{x} + \frac{1}{9} xe^x + \frac{1}{9} e^x

    then substitute x= o in last both result and solve for c_1, c_2
    Last edited by kjchauhan; December 17th 2010 at 01:23 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    These kind of problems are just routine knowing the corresponding theory. What difficulties have you had?.

    Fernando Revilla
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  4. #4
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    Apparently the answer doesn't end after finding the auxiliary eqn, the eqn still need to be solve for y''+7y'-8y = e^x, which is the particular solution of that eqn.
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  5. #5
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    Yes, and Kjchauhan did that rather "offhandedly" with " \frac{1}{D^2+7D-8} e^x= \frac{1}{9}xe^x". Quite a jump!

    You can solve for the particular solution by "variation of parameters" or, more simply in this case, "undetermined coefficients". The right hand side of the equation is e^x so normally, we would try something of the form y= Ae^x. But we have already seen that e^x was a solution to the associated homogeneous equation so there will be no value of A that will give anything but "0" on the left.

    Instead try y= Axe^x. Then y'= Ae^x+ Axe^x and y"= 2Ae^x+ Axe^x. Putting those into the equation gives y"+ 7y'- 8y= 2Ae^x+ Axe^x+ 7Ae^x+ 7Axe^x- 8Ae^x= (2Ae^x+ 7Ae^x)+ (Axe^x+ 7Axe^x- 8Axe^x)= 9Ae^x so we can take A= \frac{1}{9}.

    That gives y(x)= c_1e^x+ c_2e^{-8x}+ \frac{1}{9}xe^x as kjchauhan said.

    Of course, you are still not finished. Now find c_1 and c_2 so that y(0)= 8/9 and y'(0)= 1.
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  6. #6
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    Erm,Say for example, if the equation ends up with y''+7y'-8y = e^(-2x), Does it mean that I will assume form y to be Axe^(-2x)
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  7. #7
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    No, since \displaystyle e^{-2x} is not a term that appears in your homogeneous solution. You can just choose \displaystyle Ae^{-2x} as the nonhomogeneous solution.
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