Question

$$ y^{\prime \prime}+4 y^{\prime}+4 y=-\delta(t), \quad y(0)=1, \quad y_{-}^{\prime}(0)=5 $$

Short answer

Answer: The final solution is $$y(t) = A e^{-2t} + B t e^{-2t} + \left(1 - \frac{3}{2}t\right) e^{-2t}.$$

Step-by-step solution

1. Find the general solution of the homogeneous equation

We start by finding the general solution of the homogeneous equation: $$ y^{\prime \prime}+4 y^{\prime}+4 y=0 $$ To solve this, we assume the trial solution $$y(t) = e^{r t}$$. Plugging this into the ODE, we get the characteristic equation: $$ r^2 + 4r + 4 = 0 $$ Solving for r, we find the roots to be $$r_1 = r_2 = -2$$ as the equation is a perfect square trinomial: $$(r + 2) ^2 = 0$$. Since there's only one distinct root, the general solution for the homogeneous equation is given by: $$ y_h(t) = A e^{-2t} + B t e^{-2t} $$ where A and B are constants.

2. Finding the particular solution of the inhomogeneous equation

To find a particular solution for the inhomogeneous equation, we consider the Laplace Transform method. Taking the Laplace Transform of the given ODE, we have: $$ \mathcal{L}\{y^{\prime \prime} + 4 y^{\prime} + 4 y = -\delta(t)\} $$ Applying the properties of the Laplace Transform, we get: $$ s^2 Y(s) - sy(0) - y_{-}^{\prime}(0) + 4(sY(s) - y(0)) + 4Y(s) = - \mathcal{L}\{\delta(t)\} $$ Substitute the initial conditions $$y(0) = 1$$ and $$y_{-}^{\prime}(0) = 5$$, we obtain: $$ s^2 Y(s) - s - 5 + 4s Y(s) - 4 + 4 Y(s) = -1 $$ Rearranging and factoring out $$Y(s)$$, we have: $$ Y(s) = \frac{1+s+5}{(s^2+4s+4)} $$ Now, we will take the inverse Laplace Transform to obtain the particular solution: $$ y_p(t) = \mathcal{L}^{-1}\{Y(s)\} $$ Since the denominator is in the form of a perfect square (here, $$(s+2)^2$$), we can use the inverse Laplace table to find that the inverse Laplace Transform: $$ y_p(t) = \left(1 - \frac{3}{2}t\right) e^{-2t} $$

3. Apply initial conditions to find the overall solution

The overall solution is given by the sum of the homogeneous and particular solutions: $$ y(t) = y_h(t) + y_p(t) = A e^{-2t} + B t e^{-2t} + \left(1 - \frac{3}{2}t\right) e^{-2t} $$ The initial conditions are already used to obtain the particular solution, and the general solution contains two constants A and B. So, the final solution is given by: $$ y(t) = A e^{-2t} + B t e^{-2t} + \left(1 - \frac{3}{2}t\right) e^{-2t} $$

Key concepts

Characteristic Equation

The characteristic equation is a fundamental part of solving linear homogeneous differential equations. It is obtained by substituting a trial solution of the form \( e^{rt} \) into the homogeneous differential equation. For the given exercise, the characteristic equation is \( r^2 + 4r + 4 = 0 \), which is solved to find the roots that determine the behavior of the solution. Roots can be real and distinct, real and repeated, or complex. In this case, since the characteristic equation is a perfect square, we have a repeated root, \( r = -2 \), which signifies that the solution to the homogeneous equation involves terms that decay exponentially over time.

Homogeneous Solutions

Homogeneous solutions, also known as complementary solutions, arise from the associated homogeneous differential equation where the right-hand side is set to zero. When solving an equation like \( y'' + 4y' + 4y = 0 \), we look for solutions that satisfy the equation without the forcing term. With the repeated root \( r = -2 \), the homogeneous solution takes a specific form \( y_h(t) = Ae^{-2t} + Bte^{-2t} \), where \( A \) and \( B \) are constants to be later determined by initial conditions. This solution captures the system's response free of external influences.

Particular Solutions

Particular solutions specifically address the non-homogeneous part of a differential equation. They are found by considering the form of the non-homogeneous term, in this case \( -\text{\( \delta \)}(t) \), and applying an appropriate method to solve for it—often the method of undetermined coefficients or variation of parameters. In our exercise, we employ the Laplace Transform to find the particular solution. The Laplace Transform is powerful as it converts the differential equation into an algebraic equation, which is easier to solve. After applying initial conditions, the particular solution \( y_p(t) \) is obtained, representing the forced response of the system to external input.

Laplace Transforms

Laplace Transforms are a widely used integral transform in engineering and mathematics for solving linear differential equations. They turn complex time-domain problems into simpler s-domain algebraic equations. The transform is particularly useful for handling initial value problems with non-zero initial conditions. By taking the Laplace Transform of each term in the differential equation, we simplify the problem significantly. For inhomogeneous equations, the Laplace Transform of any forcing functions, like a Dirac delta function, can be integrated into the process. Once the algebraic equation is solved for \( Y(s) \), the inverse Laplace Transform is applied to retrieve the solution in the time domain, \( y_p(t) \).

Initial Conditions

Initial conditions are the values of the function and its derivatives at the initial time, often \( t = 0 \). They are essential for determining the specific solution of a differential equation. Given the initial conditions \( y(0) = 1 \) and \( y'_{-}(0) = 5 \), they allow us to solve for the constants in the homogeneous solution, ensuring that the solution fits the specific scenario of the problem. Additionally, these conditions help transform the Laplace domain solution back into the time domain by giving us the constants needed to complete the particular solution, ultimately leading to the overall solution that satisfies both the differential equation and the initial conditions.