Consider the second-order initial value problem y''(x)+4y'(x)+5y(x)=0 with y(0)=3,y'(0)=-5 (a) Reformulate the given second-order equation as an equivalent system of differential equations. (b) Solve the reformulated system over the interval [0,0.2] using the Kutta method (RK2) with a step size h=0.1

(a) To reformulate the given second-order equation as an equivalent system of differential equations, we can introduce a new variable $z(x)$ such that $z(x) = y'(x)$. Then, the equation becomes:<br /><br />$y''(x) = -4y'(x) - 5y(x)$<br /><br />Now, we can rewrite this equation in terms of $y'(x)$ and $y(x)$:<br /><br />$y'(x) = z(x)$<br /><br />$y''(x) = -4z(x) - 5y(x)$<br /><br />So, the equivalent system of differential equations is:<br /><br />$\begin{cases} y'(x) = z(x) \\ z'(x) = -4z(x) - 5y(x) \end{cases}$<br /><br />(b) To solve the reformulated system over the interval $[0,0.2]$ using the Kutta method (RK2) with a step size $h=0.1$, we can follow these steps:<br /><br />1. Initialize the values of $y(0)$ and $z(0)$ as $3$ and $-5$, respectively.<br />2. For each step, calculate the values of $y$ and $z$ using the Kutta method formulas:<br />$\begin{align*}<br />k_1 &= f(x_n, y_n, z_n) \\<br />k_2 &= f(x_n + \frac{h}{2}k_1, y_n + \frac{h}{2}k_1, z_n + \h}{2}k_1) \\<br />k_3 &= f(x_n + \frac{h}{2}k_2, y_n + \frac{h}{2}k_2, z_n + \frac{h}{2}k_2) \\<br />k_4 &= f(x_n + hk_3, y_n + hk_3, z_n + hk_3) \\<br />y_{n+1} &= y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \\<br />z_{n+1} &= z_n + \frac{h}{6}(k_1 + k_2 + k_3 + k_4)<br />\end{align*}$<br />where $f(x, y, z) = [-z, -4z - 5y]$.<br /><br />3. Repeat step 2 for $n = 0, 1, 2, \ldots, 19$ to obtain the values of $y$ and $z$ at each step.<br />4. The final values of $y$ and $z$ at $x = 0.2$ will be the solution to the reformulated system over the interval $[0,0.2]$.<br /><br />Note: The Kutta method formulas provided above are for the RK2 method. There are other Kutta methods (e.g., RK4) that can be used for solving differential equations, but the formulas would be different.