We are tasked with solving the second-order linear differential equation with constant coefficients:

$y'' + 4y' + 17y = 0$ with the initial conditions: $y(0) = -2, \quad y'(0) = -3.$ We need to find $y(x)$, with $x$ as the independent variable.

Step 1: Solve the characteristic equation

For the given differential equation $y'' + 4y' + 17y = 0$, the characteristic equation is:

$r^2 + 4r + 17 = 0.$

Step 2: Solve for $r$

We solve this quadratic equation using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ Here, $a = 1$, $b = 4$, and $c = 17$. Substituting these values:

$r = \frac{-4 \pm \sqrt{4^2 - 4(1)(17)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 68}}{2} = \frac{-4 \pm \sqrt{-52}}{2} = \frac{-4 \pm 2i\sqrt{13}}{2}.$

Thus, the roots are: $r = -2 \pm i\sqrt{13}.$

Step 3: General solution

The general solution for a differential equation with complex roots $r = \alpha \pm \beta i$ is given by:

$y(x) = e^{\alpha x} \left( C_1 \cos(\beta x) + C_2 \sin(\beta x) \right).$

In this case, $\alpha = -2$ and $\beta = \sqrt{13}$, so the general solution is:

$y(x) = e^{-2x} \left( C_1 \cos(\sqrt{13}x) + C_2 \sin(\sqrt{13}x) \right).$

Step 4: Apply initial conditions

We now apply the initial conditions $y(0) = -2$ and $y'(0) = -3$.

1. Apply $y(0) = -2$:

$y(0) = e^{-2(0)} \left( C_1 \cos(0) + C_2 \sin(0) \right) = C_1.$ Thus, $C_1 = -2$.

2. Apply $y'(0) = -3$:

First, find $y'(x)$: $y'(x) = \frac{d}{dx} \left[ e^{-2x} \left( C_1 \cos(\sqrt{13}x) + C_2 \sin(\sqrt{13}x) \right) \right].$ Using the product rule: $y'(x) = e^{-2x} \left( -2(C_1 \cos(\sqrt{13}x) + C_2 \sin(\sqrt{13}x)) + C_1 (-\sqrt{13} \sin(\sqrt{13}x)) + C_2 \sqrt{13} \cos(\sqrt{13}x) \right).$

At $x = 0$: $y'(0) = e^{0} \left[ -2(C_1 \cos(0) + C_2 \sin(0)) + C_1(0) + C_2 \sqrt{13} \cos(0) \right] = -2C_1 + \sqrt{13}C_2.$ Substituting $y'(0) = -3$ and $C_1 = -2$: $-2(-2) + \sqrt{13}C_2 = -3,$ $4 + \sqrt{13}C_2 = -3,$ $\sqrt{13}C_2 = -7,$ $C_2 = \frac{-7}{\sqrt{13}}.$

Step 5: Final solution

Substitute $C_1 = -2$ and $C_2 = \frac{-7}{\sqrt{13}}$ into the general solution:

$y(x) = e^{-2x} \left( -2 \cos(\sqrt{13}x) - \frac{7}{\sqrt{13}} \sin(\sqrt{13}x) \right).$

This is the solution to the initial value problem.

Would you like further details or clarification?

Here are 5 questions for further understanding:

1. What is the significance of complex roots in differential equations?
2. How does the initial condition $y'(0) = -3$ influence the solution?
3. Could you apply the method to an equation with different coefficients?
4. Why do we use Euler’s formula for solutions involving complex roots?
5. What would happen if the roots of the characteristic equation were real instead of complex?

Tip: Always check your initial conditions carefully after solving, to ensure your constants are correctly calculated.