# 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.

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

you did not use -b.

## To solve the initial-value problem, let's correct the mistake in the quadratic formula first.

The equation is given as y'' + 4y' + 6y = 0, which represents a second-order linear homogeneous differential equation. We can solve it by finding the roots of the characteristic equation, which is obtained by substituting y = e^(rx) into the differential equation.

So, let's find the roots of the characteristic equation:

The characteristic equation is r^2 + 4r + 6 = 0.

Now, we can apply the quadratic formula: r = (-b ± √(b^2 - 4ac)) / (2a).

The coefficients for the quadratic equation are a = 1, b = 4, and c = 6.

Plugging these values into the quadratic formula gives:

r = (-4 ± √(4^2 - 4(1)(6))) / (2(1)).
r = (-4 ± √(16 - 24)) / 2.
r = (-4 ± √(-8)) / 2.
r = (-4 ± 2√2i) / 2.
r = -2 ± √2i.

Now that we have the correct roots, let's proceed with finding the solution.

Using the complex roots, we have alpha = -2 and beta = √2.

The general solution of the differential equation is given by:

y(x) = e^(αx) * (c1 * cos(βx) + c2 * sin(βx)).

Now, let's apply the initial conditions to determine the specific solution.

Given y(0) = 2, we have y(0) = e^(-2*0) * (c1 * cos(√2*0) + c2 * sin(√2*0)) = c1.

So, c1 = 2.

Now, given y'(0) = 4, we can find the derivative of y(x) with respect to x and evaluate it at x = 0.

y'(x) = e^(αx) * (-α * (c1 * sin(βx) + c2 * cos(βx)) + β * (c1 * cos(βx) + c2 * sin(βx))).

Evaluating y'(x) at x = 0:

y'(0) = e^(-2*0) * (-2 * (c1 * sin(√2*0) + c2 * cos(√2*0)) + √2 * (c1 * cos(√2*0) + c2 * sin(√2*0))) = -2 * c1 + √2 * c2.

Given y'(0) = 4, we have -2 * c1 + √2 * c2 = 4.

Since c1 = 2 (from the previous condition), we can substitute it in the equation:

-2 * (2) + √2 * c2 = 4.
-4 + √2 * c2 = 4.
√2 * c2 = 8.
c2 = 8 / √2.
c2 = 4√2.

Now, substituting the values of c1 and c2 back into the general solution:

y(x) = e^(-2x) * (2 * cos(√2x) + 4√2 * sin(√2x)).

This is the correct specific solution for the given initial-value problem.