Calculus - Second Order Differential Equations

Solve the initial-value problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1)
r=(16 +/- Sqrt(-8))
r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2)
y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2
y'(0)=4, c2=4

What am I doing wrong here? This is the answer that I came up with but it is incorrect. Thanks.

your use of the quadratic formula is wrong. r= (-b +- sqrt (b^2 -4ac)/2a

you did not use -b.

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asked by COFFEE

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