# Thread: Third Order Non-Homogenous ODE

1. ## 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;

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

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

2. Originally Posted by 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
But your $\displaystyle cx$ is part of the complimentary solution. You'll need to try $\displaystyle bx^3$ only.