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Math Help - Third Order Non-Homogenous ODE

  1. #1
    Member zangestu888's Avatar
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    Third Order Non-Homogenous ODE

    Iam having trouble in finding the particular solution for this DE, thas really the most hardest thing to me always but this is what i did the DE itself is;

    y'''+2y''=32e^{2x}+24x

    If found my homogenous side which turned out to be; ( am using method of undetermined coefficients we have learned all the other easy ways thier is to do this i guess)

    y(x)=c1+c2x+c3e^{-2x} (homogeneous side), for my particular guess ;

    ae^{2x}+bx+c since bx appeard in this homogenous and the constant c also appears i multiplied them by x so my final guess was

    ae^{2x}+bx^2+cx when i equate terms and solve i get a=2 b=0 c=0? am not sure were my mistake is or if my intial guess itself is incorrect! thanks!! in advance
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  2. #2
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    Quote Originally Posted by zangestu888 View Post
    Iam having trouble in finding the particular solution for this DE, thas really the most hardest thing to me always but this is what i did the DE itself is;

    y'''+2y''=32e^{2x}+24x

    If found my homogenous side which turned out to be; ( am using method of undetermined coefficients we have learned all the other easy ways thier is to do this i guess)

    y(x)=c1+c2x+c3e^{-2x} (homogeneous side), for my particular guess ;

    ae^{2x}+bx+c since bx appeard in this homogenous and the constant c also appears i multiplied them by x so my final guess was

    ae^{2x}+bx^2+cx when i equate terms and solve i get a=2 b=0 c=0? am not sure were my mistake is or if my intial guess itself is incorrect! thanks!! in advance
    But your cx is part of the complimentary solution. You'll need to try bx^3 only.
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