finding a solution/ characteristic equation

mathstresser

Junior Member
Joined
Jan 28, 2006
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134
Find a solution to y''' -6y'' +12y' -8y=0
y(0)=-2
y'(0)=0
y''(0)=2

I have the characteristic solution of the first equation, and that is

\(\displaystyle \L\\ r^{3}-6r^{2}+12r-8=0\)

which can also be written as

\(\displaystyle \L\\ (r-2)^{3}\)

What else do I do?
 
Now, you have:

\(\displaystyle \L\\y=C_{1}e^{2x}+C_{2}xe^{2x}+C_{3}x^{2}e^{2x}\)

Find y' and y'', use your initial conditions to find \(\displaystyle \L\\C_{1}, \;\ C_{2}, \;\ C_{3}\)

I will do the first one and you finish.

Since y(0)=-2, then:

\(\displaystyle \L\\-2=C_{1}e^{2(0)}; \;\ C_{1}=-2\)
 
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