2nd order linear w/ b^2-4ac<0: y''+4y'+8y=0, y(0)=1, y'(0

warsatan

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Sep 12, 2005
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just started this chapter and this problem give me some problem,

solve y''+4y'+8y=0 with y(0)=1 and y'(0)=3.

thank you.
 
Set up your characteristic equation using the quadratic.

\(\displaystyle \L\\m^{2}+4m+8\)

\(\displaystyle m_{1}={\alpha}+i{\beta} \;\ and \;\ m_{2}={\alpha}-i{\beta}\)

Use \(\displaystyle \L\\y=e^{{\alpha}x}\left(C_{1}cos({\beta}x)+C_{2}sin({\beta}x)\right)\)

This has only complex roots: \(\displaystyle \L\\-2+2i \;\ and \;\ -2-2i\)

\(\displaystyle y=e^{-2x}\left(C_{1}cos(2x)+C_{2}sin(2x)\right)\)

Now, using your initial conditions, can you find C1 and C2?.
 
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