Calculus
posted by COFFEE .
Please look at my work below:
Solve the initialvalue problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r^2+4r+6=0,
r=(16 +/ Sqrt(4^24(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
y(x)=e^(8x)*(2*cos(Sqrt(2)x)+4*sin(Sqrt(2)x))
What am I doing wrong here? This is the answer that I came up with but it is incorrect. Thanks.
For Further Reading
* Calculus  Second Order Differential Equations  bobpursley, Monday, July 9, 2007 at 10:09pm
your use of the quadratic formula is wrong. r= (b + sqrt (b^2 4ac)/2a
you did not use b.

y''+4y'+6y=0, y(0)=2, y'(0)=4
r^2+4r+6=0, r=(4 +/ sqrt(164(1)(6))/2
r=2 +/ sqrt(2)*i
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)(01/sqrt(2)*c2)=4
c2=2/sqrt(2)
y(x)=e^2x*(2cos(sqrt(2))x+(2/sqrt(2))sin(sqrt(x))x)
What is wrong with my solution? thanks.
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