# Third Order Non-Homogenous ODE

• Jun 24th 2009, 06:46 PM
zangestu888
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
• Jun 25th 2009, 04:08 AM
Jester
Quote:

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.