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**zangestu888** 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;

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

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)

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

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

$\displaystyle ae^{2x}+bx^2+cx$ 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