Wednesday
July 23, 2014

Homework Help: Math/Calculus

Posted by COFFEE on Thursday, July 12, 2007 at 12:15am.

Solve the initial-value problem.

Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks.

y''+4y'+6y=0, y(0)=2, y'(0)=4
r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)(6))/2
r=-2 +/- sqrt(2)*i , alpha = -2, beta = 2(sqrt(2))

y=e^-2x*(c1*cos(sqrt(2))x+c2*sin(sqrt(2))x)
y(0)=1*(c1+0)=2, c1=2
y'=(-1/2)e^-2x*(c1*(sin(sqrt(2)))/sqrt(2)-c2*(cos(sqrt(2)))/sqrt(2))
y'(0)=(-1/2)(0-1/sqrt(2)*c2)=4
c2=2/sqrt(2)
y(x)=e^-2x*(2cos(sqrt(2))x+(2/sqrt(2))sin(sqrt(x))x)

I dont follow the y', recheck it. How did you get the 1/2 coefficient?

I got it...thanks.

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