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 two first-order differential equations. (b) Solve the reformulated system over the interval [0,0.2] using the second-order Runge- Kutta method (RK2) with a step size h=0.1

(a) To reformulate the given second-order equation as an equivalent system of two first-order differential equations, we introduce a new variable $z(x)$ such that $z(x) = y'(x)$. Then, we can rewrite the given equation as:<br /><br />$y'(x) = z(x)$<br />$y''(x) = z'(x)$<br /><br />Substituting these into the original equation, we get:<br /><br />$z'(x) + 4z(x) + 5y(x) = 0$<br /><br />Now, we have a system of two first-order differential equations:<br /><br />$y'(x) = z(x)$<br />$z'(x) = -4z(x) - 5y(x)$<br /><br />(b) To solve the reformulated system over the interval $[0,0.2]$ using the second-order Runge-Kutta method $(RK2)$ with a step size $h=0.1$, we can follow these steps:<br /><br />1. Initialize the values: $x_0 = 0$, $y_0 = 3$, $z_0 = -5$, and $h = 0.1$.<br />2. For $i = 1, 2,..., N$, where $N = \frac{0.2}{h} = 2$, do the following:<br /> a. Calculate the intermediate values:<br /> $k_1y = y_{i-1} + h \cdot f(x_{i-1}, y_{i-1}, z_{i-1})$<br /> $k_1z = z_{i-1} + h \cdot g(x_{i-1}, y_{i-1}, z_{i-1})$<br /> b. Update the values:<br /> $y_i = y_{i-1} + \frac{h}{2} \cdot (k_1y + k_2y)$<br /> $z_i = z_{i-1} + \frac{h}{2} \cdot (k_1z + k_2z)$<br /> c. Calculate the error:<br /> $e_y = y_i - y_{i-1}$<br /> $e_z = z_i - z_{i-1}$<br /> d. Update the values:<br /> $y_i = y_i + \frac{e_y}{1 + e_y + e_z}$<br /> $z_i = z_i + \frac{e_z}{1 + e_y + e_z}$<br /> e. Update the value of $x$: $x_i = x_{i-1} + h$<br /><br />After performing these steps for $i = 1, 2,..., N$, we will have the approximate solutions for $y(x)$ and $z(x)$ over the interval $[0,0.2]$ using the second-order Runge-Kutta method $(RK2)$ with a step size $h=0.1$.