91Ó°ÊÓ

First, we need to solve the homogeneous equation: \[ y^{\prime \prime} - 2 y^{\prime} + 10y = 0 \] The characteristic equation for this differential equation is \[ r^2 - 2r + 10 = 0 \] Solving the above quadratic equation using the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get complex conjugate roots \[ r_{1,2} = \frac{2 \pm i \sqrt{32}}{2} = 1 \pm 4i \] Thus, the complementary solution is given by \[ y_c(t) = C_1e^{t} \cos(4t) + C_2 e^{t} \sin(4t) \]

Next, we will find the particular solution. Since the forcing term is \(\cos(2t)\), we will assume a particular solution of the form \[ y_p(t) = A\cos(2t) + B\sin(2t) \] Now, we need to find the first and second derivatives of \(y_p(t)\): \[ y^{\prime}_p(t) = -2A\sin(2t) + 2B\cos(2t) \] \[ y^{\prime\prime}_p(t) = -4A\cos(2t) - 4B\sin(2t) \] Substitute \(y_p(t), y^{\prime}_p(t),\) and \(y^{\prime\prime}_p(t)\) into the given non-homogeneous equation: \[ -4A\cos(2t) - 4B\sin(2t) - 2(-2A\sin(2t) + 2B\cos(2t)) + 10( A\cos(2t) + B \sin(2t) ) = \cos(2t) \] Now, combining like terms, we have: \[ (-4A + 4B + 10A)\cos(2t) + (-4B - 4A + 10B)\sin(2t) = \cos(2t) \] Equating the coefficients, we get the following system of equations: \[ 6A + 4B = 1 \] \[ -4A + 6B = 0 \] Solving the system, we find \(A = \frac{1}{10}\) and \(B = \frac{1}{5}\). Thus, the particular solution is: \[ y_p(t) = \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \]

Now we need to apply the initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 1\) to find the values of \(C_1\) and \(C_2\). The general solution is the sum of the complementary and particular solutions: \[ y(t) = y_c(t) + y_p(t) = C_1e^{t}\cos(4t) + C_2 e^{t}\sin(4t) + \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \] Applying the first initial condition \(y(0) = 0\), we get: \[ 0 = C_1 + \frac{1}{10} \] So, \(C_1 = -\frac{1}{10}\). Now we need to apply the second initial condition, \(y^{\prime}(0) = 1\). First, let's find the derivative of our general solution: \[ y^{\prime}(t) = -\frac{1}{10}e^{t}\cos(4t) - 4 C_2 e^{t}\cos(4t) - 2 \frac{1}{10}\sin(2t) + 2 \frac{1}{5}\cos(2t) \] Applying the condition \(y^{\prime}(0) = 1\), we have: \[ 1 = -\frac{1}{10} -\frac{1}{5} => C_2 = \frac{1}{2} \] The values of \(C_1\) and \(C_2\) are now determined, and the particular solution is: \[ y(t) = -\frac{1}{10}e^{t}\cos(4t) + \frac{1}{2} e^{t}\sin(4t) + \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \] Now we have found the required solution for the given initial value problem, and further evaluation using convolution integral will not be performed as per the exercise prompt.