x''+x'+2x=0 x(0)=2 x'(0)=0(adsbygoogle = window.adsbygoogle || []).push({});

I've taken the characteristic equation and reduced the roots to

1/2 +- Sqrt(7/4)i of the form

a +- bi (i = sqrt(-1)

Then i put the homogeneous solution into the form of e[tex]^{}at[/tex]*(B1cos(bt)+B2sin(bt))

for B1 i used the first i.c. and found that B1=2

for B2 i used the i.c. for x' and found B2 to be 1/(sqrt7/4)

now i need to get it into the form of Ae[tex]^{}at[/tex]sin(bt+phi)

where A = sqrt(B1^2+B2^2) and phi is acos(B1/A)

the problem is that A is not the hypotenuse of of a right triangle. I didn't know if i screwed up my b values along the way or if i should go to the law of cosines, but i saw no mention of such a problem so I figured i may have made an error beforehand. any help is appreciated.

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# Homework Help: 2nd order ode w/complex conjugate roots

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