Calculus  Second Order Differential Equations
posted by COFFEE .
Solve the initialvalue problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r^2+4r+6=0,
r=(16 +/ Sqrt(4^24(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.
Respond to this Question
Similar Questions

Inequality
When I solve the inquality 2x^2  6 < 0, I get x < + or  sqrt(3) So how do I write the solution? 
Math(Roots)
sqrt(24) *I don't really get this stuff.Can somebody please help me? 
math,algebra,help
Directions are simplify by combining like terms. x radiacal 18 3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which reduces … 
Mathematics
sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So you can see the steps — sqrt 6 * sqrt 8 = sqrt 48 sqrt 7 * sqrt 5 = sqrt 35 I hope this helps a little more. Thanks for asking. 
Math Help please!!
Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine … 
Calculus  Second Order Differential Equations
Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 … 
Calculus
Please look at my work below: Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(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 … 
Math/Calculus
Solve the initialvalue problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? 
maths
Choose the option that gives the solution of the initialvalue problem dy dx=(1 + 2 cos2x^2)/y (y > 0), y= 1 when x = 0. Options A y = 1+ 2sin^2 x B y = (1 + 2 sin x)^2 C y = (4x + cos(2x))^2 D y = 4x + cos^2 x E y =sqrt(4x + cos(2x) … 
Calculus
In these complex exponential problems, solve for x: 1)e^(i*pi) + 2e^(i*pi/4)=?