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.

## Answer This Question

## Related Questions

- Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...
- Calculus - Second Order Differential Equations - Solve the initial-value problem...
- Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...
- calc check: curve length - Find the length of the curve y=(1/(x^2)) from ( 1, 1...
- calculus - how do you solve the initial value problem by using separation of ...
- Algebra - sqrts - Square roots. Woohoo. Want to check some work I did. 1. ...
- math - how would you simplify this equation: y = (x+3)/[(4-sqrt(16+h))] please ...
- Math - So I am supposed to solve this without using a calculator: Sqrt[20]/10 - ...
- Math Help please!! - Could someone show me how to solve these problems step by ...
- Advanced Math - Can someone check these for me? Please? Use half-angle identity ...

More Related Questions