Question

Solve the initial value problem. Find a formula that does not involve step functions and represents \(y\) on each subinterval of \([0, \infty)\) on which the forcing function is zero. (a) \(y^{\prime \prime}-y=\sum_{k=1}^{\infty} \delta(t-k), \quad y(0)=0, \quad y^{\prime}(0)=1\) (b) \(y^{\prime \prime}+y=\sum_{k=1}^{\infty} \delta(t-2 k \pi), \quad y(0)=0, \quad y^{\prime}(0)=1\) (c) \(y^{\prime \prime}-3 y^{\prime}+2 y=\sum_{k=1}^{\infty} \delta(t-k), \quad y(0)=0, \quad y^{\prime}(0)=1\) (d) \(y^{\prime \prime}+y=\sum_{k=1}^{\infty} \delta(t-k \pi), \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

Question: Find the formula for y(t) that does not involve step functions, representing the solution for the differential equation problem (a) on each subinterval of [0, ∞) where the forcing function is zero. Answer: For problem (a), the formula for y(t) is as follows: For \(0 \le t < 1\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t}\) For \(1 \le t < 2\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t} + e^{-(t-1)}\) For \(2 \le t < 3\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t} + e^{-(t-1)} + e^{-(t-2)}\) And so on, with each subinterval having a similar structure as the pattern continues.

Step-by-step solution

Find homogeneous solution

Find the homogeneous solution to the given differential equation \(y^{\prime \prime}-y=0\) by solving its characteristic equation, which is \(r^2-1=0\).

Find particular solution

Find the particular solution to the differential equation by using the Laplace Transform method and then applying the Inverse Laplace Transform to obtain the particular solution in terms of \(y(t)\).

Combine homogeneous and particular solutions

Combine the homogeneous and particular solutions to build the general solution to the differential equation.

Apply initial conditions

Apply the given initial conditions \(y(0)=0\) and \(y^{\prime}(0)=1\) to find the specific values of the constants in the general solution.

Write the final formula

Write the final formula for \(y(t)\) that does not involve step functions and represents the solution on each subinterval where the forcing function is zero. Now let's apply these steps to problem (a):

Find homogeneous solution

Our characteristic equation is \(r^2-1=0\). Solving gives us the roots \(r=1\) and \(r=-1\). Therefore, the homogeneous solution is: \(y_h(t) = C_1 e^t + C_2 e^{-t}\)

Find particular solution

Let \(G(t)=\sum_{k=1}^{\infty} \delta(t-k)\). Taking the Laplace Transform of the given differential equation, we get: \(Y(s) (s^2 - 1) - s Y(0^+) - Y^{\prime}(0^+) = G(s)\) Using the initial conditions \(y(0)=0\) and \(y^{\prime}(0)=1\), we get: \(Y(s) (s^2 - 1) - s (0) - 1 = G(s)\) Now taking the Inverse Laplace Transform, we find the particular solution: \(y_p(t) = \sum_{k=1}^{\infty} e^{-(t-k)} u(t-k)\)

Combine homogeneous and particular solutions

The general solution is the sum of the homogeneous and particular solutions: \(y(t) = y_h(t) + y_p(t)\) \(y(t) = C_1 e^t + C_2 e^{-t} + \sum_{k=1}^{\infty} e^{-(t-k)} u(t-k)\)

Apply initial conditions

Applying the initial condition \(y(0)=0\), we get: \(0 = C_1 e^0 + C_2 e^0 + 0\) \(0 = C_1 + C_2\) Applying the initial condition \(y^{\prime}(0)=1\) gives us: \(1 = C_1 e^0 - C_2 e^0\) \(1 = C_1 - C_2\) Solving these equations simultaneously, we obtain \(C_1=\frac{1}{2}\) and \(C_2=-\frac{1}{2}\).

Write the final formula

Substituting the values of the constants back into the general solution, we get: \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t} + \sum_{k=1}^{\infty} e^{-(t-k)} u(t-k)\) To eliminate the step functions in the solution, break it down into subintervals where the forcing function is zero. Between every two natural numbers \((k-1)\) and \(k\), the forcing function is zero. Thus, we have: For \(0 \le t < 1\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t}\) For \(1 \le t < 2\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t} + e^{-(t-1)}\) For \(2 \le t < 3\): \(y(t) = \frac{1}{2} e^t - \frac{1}{2} e^{-t} + e^{-(t-1)} + e^{-(t-2)}\) And so on. Now apply a similar methodology to solve problems (b), (c), and (d).

Key concepts

Initial Value Problem

An initial value problem involves solving a differential equation with given initial conditions. These conditions specify the value of the solution and its derivatives at a starting point, usually when time \( t = 0 \). For example, in our exercise, \( y(0) = 0 \) and \( y'(0) = 1 \) are the initial conditions.

This type of problem helps us determine a unique solution for the differential equation. The initial conditions allow us to solve for any arbitrary constants in the solution, making the result specific to the problem at hand.

Homogeneous Solution

A homogeneous solution is part of solving a differential equation where we only consider the equation set to zero. In the exercise, the homogeneous equation is \( y'' - y = 0 \). The characteristic equation, \( r^2 - 1 = 0 \), is derived by supposing a solution of a certain form, usually \( e^{rt} \).

Solving this characteristic equation gives the roots, which in this case are \( r = 1 \) and \( r = -1 \). These roots define the homogeneous solution as:
\[ y_h(t) = C_1 e^t + C_2 e^{-t} \]
This part of the solution doesn’t account for any forcing function (external input), and the constants \( C_1 \) and \( C_2 \) will later be determined by the initial conditions.

Laplace Transform

The Laplace Transform is a powerful tool for solving differential equations, particularly with initial value problems. It transforms a differential equation in the time domain into an algebraic equation in the \( s \)-domain.

In our problem, transforming \( y'' - y = \sum_{k=1}^{\infty} \delta(t-k) \) gives:
\( Y(s)(s^2 - 1) - sY(0^+) - Y'(0^+) = G(s) \)
Here, \( Y(s) \) represents the transformed function, making it easier to handle, especially with conditions or inputs like delta functions.

Once transformed, solving equations becomes more straightforward, and inverse transforming the result returns us back to the time domain.

Inverse Laplace Transform

The Inverse Laplace Transform helps us convert the solution back from the \( s \)-domain to the time domain. After solving the algebraic equation in the \( s \)-domain, we need to find the corresponding time-domain solution.

Using tables or specific techniques, like partial fraction decomposition, assists in finding the inverse transformation. In our case, by applying the inverse transform, we find:

\( y_p(t) = \sum_{k=1}^{\infty} e^{-(t-k)} u(t-k) \)
Where \( y_p(t) \) is a superposition of responses to each impulse in the forcing function. This step is crucial for interpreting the final solution in terms we can understand and work with in time.

Forcing Function

A forcing function represents an external input to the system expressed by a differential equation. It can be anything that drives the system, like a delta function impulse.

In this exercise, the forcing function is \( \sum_{k=1}^{\infty} \delta(t-k) \), representing an infinite series of impulses at each integer time \( t = k \). These delta functions activate the system's response at specific times.

The methodical handling of these impulses via Laplace Transform means each impulse is tackled individually, making understanding each part of the response simpler. Breaking down the interval where the forcing function is zero allows finding solutions segment by segment, ensuring our final solution accounts precisely for each interval.