# 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.

## In your first attempt, you made a mistake when using the quadratic formula. The correct form of the quadratic formula is r = (-b ± sqrt(b^2 - 4ac)) / (2a). However, you forgot to include the -b term in the formula. As a result, you obtained incorrect values for r.

In the corrected solution, you correctly applied the quadratic formula to find the values of r. You obtained r = -2 ± sqrt(2)i. However, there are a few errors in the subsequent steps of your solution.

Firstly, when solving for the constant c1 at y(0), you incorrectly set y(0) equal to c1. It should be e^(8*0) * (c1*cos(0) + c2*sin(0)) = 2. Since cosine(0) = 1 and sine(0) = 0, this simplifies to e^0 * (c1*1 + c2*0) = 2, which gives c1 = 2.

Secondly, when solving for the constant c2 at y'(0), you made an error in your calculations. The correct expression for y'(0) is given by e^(8*0) * (-2*c1*sin(sqrt(2)*0) + c2*cos(sqrt(2)*0)) = 4. Simplifying this, we have -2*c1 + c2 = 4. Since c1 = 2 (from the previous step), we can substitute this value to find c2 = 2/sqrt(2).

Finally, the correct solution to the initial-value problem is y(x) = e^(-2x) * (2*cos(sqrt(2)*x) + (2/sqrt(2))*sin(sqrt(2)*x)), which matches your final answer.

So, the mistake in your solution was the incorrect use of the quadratic formula and the errors in obtaining the values of c1 and c2. By correcting these mistakes, you arrived at the correct solution.