Thursday

December 18, 2014

December 18, 2014

Posted by **COFFEE** on Thursday, July 12, 2007 at 12:15am.

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

Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...

Math - Consider the initial value problem y'' +5y'+6y=0, y(0)=4.87 and y'(0)=...

Math - A series circuit contains a resistor with R = 24 , an inductor with L = 2...

Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...

College Finite Math - Suppose for this problem that Pr[E]=13/24 and Pr[F]=5/8 ...

calculus - how do you solve the initial value problem by using separation of ...

calculus - how do you solve the initial value problem by using separation of ...

calculus - how do you solve the initial value problem by using separation of ...

Calculus II - Solve the initial value problem using Taylor Series and the ...