Thursday

July 24, 2014

July 24, 2014

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

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.

**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'' + ...

Calculus - Second Order Differential Equations - Solve the boundary-value ...

Calculus - Second Order Differential Equations - Solve the initial-value problem...

Differential Equations (Another) Cont. - For the following initial value problem...

Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...

Calculus - solve second-order initial value problem y"(x)= x e^-2x y'(0)= 0 y(0...

Differential Equations - Use the Laplace Transform to solve this initial value ...

Calculus - Solve the differential and initial value problem: x(dy/dx) + 1 = y^2 ...

Calculus - I'm having trouble with part of a question in a problem set. The ...