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

1. ## 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

2. 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 $\displaystyle Ax^3e^{-2x}$.

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### y"' - 6y" 12y' - 8y= âˆš2x e2x

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