Posted by COFFEE on Monday, July 9, 2007 at 9:10pm.
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Solve the initial-value problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r^2+4r+6=0,
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
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.
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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
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)
What is wrong with my solution? thanks.
0 answers