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

**mathwizard** · June 15th 2008, 09:58 AM

$y’’’ + 6y’’ + 12y’+ 8y = 12e^{-2x}$

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’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)? - Yahoo! Answers

**ThePerfectHacker** · June 15th 2008, 10:09 AM

Originally Posted by mathwizard

y’’’ + 6y’’ + 12y’+ 8y = 12e^(-2x)

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

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