Question

karton · Accepted Answer

We are given the following second-order linear differential equation with constant coefficients:

y^{″} - 2 y^{'} - 2 y = 3 δ (t - 1) + e^{t}

where

δ (t)

is the Dirac delta function. The initial conditions are:

y (0) = 0, y^{'} (0) = 3

To solve this differential equation, we will use the Laplace transform. Taking the Laplace transform of both sides of the equation, we get:

s^{2} Y (s) - s y (0) - y^{'} (0) - 2 s Y (s) + 2 y (0) - 2 Y (s) = 3 e^{- s} e^{s} + \frac{1}{s - 1}

Substituting the initial conditions, we get:

s^{2} Y (s) - 3 s - 2 s Y (s) = 3 + \frac{1}{s - 1}

Simplifying and solving for

Y (s)

, we get:

Y (s) = \frac{3}{s (s + 2) (s - 1)} + \frac{1}{(s - 1) (s + 2)}

We can rewrite the first term using partial fraction decomposition:

\frac{3}{s (s + 2) (s - 1)} = \frac{1}{s} - \frac{1}{s + 2} - \frac{1}{s - 1}

Substituting this into

Y (s)

, we get:

Y (s) = \frac{1}{s} - \frac{1}{s + 2} - \frac{1}{s - 1} + \frac{1}{(s - 1) (s + 2)}

Taking the inverse Laplace transform, we get the solution to the differential equation:

y (t) = ℒ^{- 1} {\frac{1}{s} - \frac{1}{s + 2} - \frac{1}{s - 1} + \frac{1}{(s - 1) (s + 2)}}

Using the properties of the Laplace transform and the inverse Laplace transform, we can simplify this expression as follows:

y (t) = 1 - e^{2 t} + e^{t} - 3 e^{t - 1} u (t - 1)

where

u (t)

is the unit step function. Therefore, the solution to the initial value problem is:

y (t) = 1 - e^{2 t} + e^{t} - 3 e^{t - 1} u (t - 1), t \geq 0

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