Calculus

posted by .

Please look at my work below:

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.

For Further Reading

* Calculus - Second Order Differential Equations - bobpursley, Monday, July 9, 2007 at 10:09pm

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

you did not use -b.
--------------------------
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
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)

What is wrong with my solution? thanks.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. math calculus please help!

    l = lim as x approaches 0 of x/(the square root of (1+x) - the square root of (1-x) decide whether: l=-1 or l=0 or l=1 Let me make sure I understand the question. Do we have lim x->0 x/[sqrt(1+x) - sqrt(1-x)] ?
  2. Inequality

    When I solve the inquality 2x^2 - 6 < 0, I get x < + or - sqrt(3) So how do I write the solution?
  3. Math(Roots)

    sqrt(24) *I don't really get this stuff.Can somebody please help me?
  4. Math

    How do you find a square root of a number that's not a perfect square?
  5. 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.
  6. 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 …
  7. Calculus - Second Order Differential Equations

    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, …
  8. Calculus - Second Order Differential Equations

    Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix 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 …
  9. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
  10. Algebra

    Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work. sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2) x*sqrt 7x-49*x ^^^ would this be my final answer?

More Similar Questions