# Differential equation y’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)

• Jun 15th 2008, 09:58 AM
mathwizard
Differential equation y’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)
\$\displaystyle y’’’ + 6y’’ + 12y’+ 8y = 12e^{-2x}\$

EDIT: Hey, why don't the primes show up? Here's it without using the code
y’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)

trying using both method of undetermined coefficient & variation of parameters.

Y&rsquo;&rsquo;&rsquo; + 6y&rsquo;&rsquo; + 12y&rsquo;+ 8y = 12e^(-2x)? - Yahoo! Answers
• Jun 15th 2008, 10:09 AM
ThePerfectHacker
Quote:

Originally Posted by mathwizard
y’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)

The first step is to solve \$\displaystyle y'''+6y''+12y'+8y = 0\$ the charachteristic equation is \$\displaystyle k^3 + 6k^2 + 12k+8 = 0\$. This factors as \$\displaystyle (k+2)^{2} = 0\$. Thus, the general solution is \$\displaystyle c_1e^{-2x}+c_2xe^{-2x}+c_3x^2e^{-2x}\$. To find the particular solution we would look for a solution of the form \$\displaystyle Ae^{-2x}\$ but since this is included amongst the general we need to look for a solution of the form \$\displaystyle Ax^3e^{-2x}\$.