Solve the initial-value problem.

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 don't follow the y', recheck it. How did you get the 1/2 coefficient?

I got it...thanks.

To solve the initial value problem, we first need to find the roots of the characteristic equation. The characteristic equation for the given differential equation is r^2 + 4r + 6 = 0.

Using the quadratic formula, we can find the roots of the equation:
r = (-4 ± √(16 - 4(1)(6))) / 2
r = (-4 ± √(16 - 24)) / 2
r = (-4 ± √(-8)) / 2
r = (-4 ± 2√2i) / 2

So, the roots of the characteristic equation are -2 ± √2i.

Now, we can write the general solution to the differential equation as:
y(x) = e^(-2x)(c1*cos(√2x) + c2*sin(√2x))

To find the values of c1 and c2, we can use the initial conditions. Given that y(0) = 2 and y'(0) = 4, we can substitute these values into the general solution and solve for c1 and c2.

Substituting x = 0 into the general solution:
y(0) = c1*cos(0) + c2*sin(0)
2 = c1

Now, let's find y'(x) using the general solution:
y'(x) = -2e^(-2x)(c1*cos(√2x) + c2*sin(√2x)) + e^(-2x)(-√2c1*sin(√2x) + √2c2*cos(√2x))

Substituting x = 0 into the expression for y'(x):
y'(0) = -2e^(-2*0)(c1*cos(√2*0) + c2*sin(√2*0)) + e^(-2*0)(-√2c1*sin(√2*0) + √2c2*cos(√2*0))
y'(0) = -2(c1) + √2c2

Given that y'(0) = 4:
4 = -2(c1) + √2c2

Solving the system of equations, we can find the values of c1 and c2. From the earlier substitution, we know that c1 = 2. Substituting this into the equation:
4 = -2(2) + √2c2
4 = -4 + √2c2
√2c2 = 8
c2 = 8 / √2
c2 = 4√2

Now, substituting the values of c1 and c2 back into the general solution, we get the particular solution:
y(x) = e^(-2x)(2*cos(√2x) + (4√2)*sin(√2x))

So, the solution to the initial value problem is y(x) = e^(-2x)(2*cos(√2x) + (4√2)*sin(√2x)).

It seems there was a mistake in the coefficients in the calculation of y'(0). The correct calculation leads to c2 = 4√2, rather than c2 = 2/√2.

To solve the initial-value problem, we start by finding the roots of the characteristic equation r^2 + 4r + 6 = 0.

Using the quadratic formula, we have r = (-4 ± sqrt(16 - 4*1*6)) / 2.
Simplifying this expression, we get r = -2 ± sqrt(2)*i.

Since the roots are complex, the general solution will be in the form y(x) = e^(-2x)(c1*cos(sqrt(2)x) + c2*sin(sqrt(2)x)).

To find the values of c1 and c2, we can use the initial conditions y(0) = 2 and y'(0) = 4:

Plugging in x = 0, we have y(0) = e^(-2*0)(c1*cos(sqrt(2)*0) + c2*sin(sqrt(2)*0)) = c1.
Therefore, c1 = 2.

Differentiating y(x) with respect to x, we get y'(x) = -2e^(-2x)(c1*sin(sqrt(2)x) + c2*cos(sqrt(2)x)) + e^(-2x)sqrt(2)(-c1*sin(sqrt(2)x) + c2*cos(sqrt(2)x)).
Simplifying this expression, we get y'(x) = e^(-2x)(sqrt(2)*c1*cos(sqrt(2)x) + (2-c2)*sin(sqrt(2)x)).

Now, let's apply the second initial condition:

Plugging in x = 0, we have y'(0) = e^(-2*0)(sqrt(2)*c1*cos(sqrt(2)*0) + (2-c2)*sin(sqrt(2)*0)) = e^0(sqrt(2)*2*1 + (2-c2)*0) = sqrt(2)*2 = 4.
Dividing both sides by sqrt(2), we get 2 = c1.

Therefore, c1 = 2.

The final solution to the initial-value problem is y(x) = e^(-2x)(2*cos(sqrt(2)x) + (2/sqrt(2))*sin(sqrt(2)x)).

Apologies for the confusion in the previous response. Please let me know if you have any further questions!