Posted by **COFFEE** on Monday, July 9, 2007 at 9:10pm.

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.

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

you did not use -b.

## Answer This Question

## Related Questions

- Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...
- Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...
- Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...
- Math Help please!! - Could someone show me how to solve these problems step by ...
- Inequality - When I solve the inquality 2x^2 - 6 < 0, I get x < + or - ...
- math,algebra,help - Directions are simplify by combining like terms. x radiacal ...
- math calculus please help! - l = lim as x approaches 0 of x/(the square root of...
- Mathematics - sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So...
- Calculus 2 (Differential Equation) - How would you solve the following problem ...
- Math(Roots) - sqrt(24) *I don't really get this stuff.Can somebody please help ...

More Related Questions