91Ó°ÊÓ

To find the complementary function, consider the homogeneous part of the given equation: \[ y'' + 4y' + 3y = 0. \] We will solve it using the characteristic equation: \[ r^2 + 4r + 3 = 0. \] Factoring the quadratic yields: \[ (r+3)(r+1) = 0. \] The roots are -3 and -1. Thus, the complementary function is given by: \[ y_c(t) = C_1 e^{-3t} + C_2 e^{-t}. \]

Now, we need to find a particular solution for the inhomogeneous equation: \[ y''+4y'+3y=\delta(t-2). \] Since the forcing term is a Dirac Delta function, it means that the system experiences an impulse at t=2. To account for this, we can write the particular solution in terms of a Heaviside step function, H(t-2), which is 0 when t<2, and 1 when t≥2. Let the particular solution be: \[ y_p(t) =C_3 e^{-3t} H(t-2) + C_4 e^{-t} H(t-2). \] Now, we will differentiate \(y_p(t)\) twice to substitute in the original equation. First derivative of \(y_p(t)\): \[ y'_p(t)=-3C_3 e^{-3t} H(t-2) - C_4 e^{-t} H(t-2) + [\delta(t-2) (C_3 e^{-6}+C_4 e^{-2})]. \] Second derivative of \(y_p(t)\): \[ y''_p(t)=9C_3 e^{-3t} H(t-2) + e^{-t} H(t-2) + \delta'(t-2) (C_3 e^{-6}+C_4 e^{-2})- 3\delta(t-2) C_3 e^{-6} - \delta(t-2) C_4 e^{-2}. \] Substituting the second derivative and the original function into the given equation, we cancel out the delta function and its first derivative terms, leaving us with: \[ 9C_3 e^{-3t} + e^{-t} - 3C_3 e^{-6} - C_4 e^{-2}=0. \] However, this equation holds only for t>2 as the equation wasn't valid for t<2. Thus, we can plug in t=2 and solve the equation. \[ 9C_3 e^{-6} + e^{-2} - 3C_3 e^{-6} - C_4 e^{-2}=0. \] Which simplifies to: \[ 6C_3 + 1 - C_4 = 0 \iff C_4 = 6C_3 + 1. \]

We are given the initial conditions: \(y(0) = 1\) and \(y'(0) = -1\). To apply these conditions, we will express the general solution in the form \(y(t) = y_c(t) + y_p(t)\) and differentiate it. Using the general solution: \[ y(t) = (C_1 e^{-3t} + C_2 e^{-t})+ (C_3 e^{-3t} H(t-2) + C_4 e^{-t} H(t-2)). \] At t=0, \[ 1= y(0) =C_1 + C_2. \] Now we differentiate y(t), \[ y'(t)= -3C_1 e^{-3t} - C_2 e^{-t} - 3C_3 e^{-3t} H(t-2) - C_4 e^{-t} H(t-2) + [\delta(t-2) (C_3 e^{-6}+C_4 e^{-2})]. \] Applying the second initial condition: \[ -1 = y'(0) = -3C_1 - C_2. \] We have two equations for the constants C1 and C2: \[ C_1 + C_2 = 1, \] \[ -3C_1 - C_2 = -1. \] Solving them simultaneously, we get: \[ C_1 = \frac{1}{2}, \quad C_2 = \frac{1}{2}. \] Now, we remember that we found \(C_4 = 6C_3 + 1\) when solving the particular solution above. However, since for \(t < 2\) there is no influence from the delta function, \(C_3\) and \(C_4\) do not change the complementary function. Hence, we will only keep the complementary function plus the particular solution we had initially:

Combining the complementary function and particular solution, we have the final general solution: \[ y(t) = \frac{1}{2} e^{-3t} + \frac{1}{2} e^{-t} + [\left(\frac{1}{2} e^{-3(t-2)}-\frac{7}{2}e^{-t}\right)H(t-2)]. \]