91Ó°ÊÓ

The homogeneous equation is given by: $$ y^{\prime\prime} - 4y = 0 $$ The characteristic equation is: $$ r^2 - 4 = 0 $$ Solving for r, we get \(r_{1}=2\) and \(r_{2}=-2\). Thus, the homogeneous solution is given by: $$ y_h(t) = C_1 e^{2t} + C_2 e^{-2t} $$

To find the particular solution, we consider the equation: $$ y^{\prime\prime} - 4y = 3 \delta(t) $$ Since the equation involves a Dirac delta function, we need to use Laplace Transforms to find the particular solution. The Laplace Transform of the equation is: $$ s^2 Y(s) - s y(0) - y'(0) - 4 Y(s) = \frac{3}{s} $$ Substituting the given initial conditions \(y(0) = -1\) and \(y'(0) = 7\), we have: $$ s^2 Y(s) + s - 7 - 4 Y(s) = \frac{3}{s} $$ Now, solve for \(Y(s)\): $$ Y(s) = \frac{3+s^2-s}{s(s^2-4)} $$ To find the inverse Laplace Transform, we perform partial fraction decomposition. The decomposed form of Y(s) is: $$ Y(s) = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s-2} $$ Solving for A, B, and C, we get \(A=\frac{3}{4} \, , \, B = -1\, , \, C = 1 \). Thus, the particular solution in the time domain is: $$ y_p(t) = \frac{3}{4} - e^{-2t} + e^{2t} $$

The general solution is the sum of the homogeneous and particular solutions: $$ y(t) = y_h(t) + y_p(t) = C_1 e^{2t} + C_2 e^{-2t} + \left(\frac{3}{4}-e^{-2t} + e^{2t}\right) $$

Since the initial conditions were used for the Laplace Transform, we don't need to solve for C1 and C2. The solution to the given differential equation is: $$ y(t) = C_1 e^{2t} + C_2 e^{-2t} + \left(\frac{3}{4} - e^{-2t} + e^{2t}\right) $$