Question

Solve the given initial value problem, in which inputs of large amplitude and short duration have been idealized as delta functions. Graph the solution that you obtain on the indicated interval. \(y^{\prime \prime}+2 y^{\prime}+y=\delta(t-2), \quad y(0)=0, \quad y^{\prime}(0)=1, \quad 0 \leq t \leq 6\)

Short answer

## Short Answer The final solution of the given initial value problem for a second-order linear ODE with a delta function is: $$ y(t) = \begin{cases} -\frac{1}{3}e^2te^{-t} & \text{for } 0 \leq t < 2\\\\ -\frac{1}{3}e^2e^{-t} + (\frac{2}{3}e^2 - 1)te^{-t} & \quad \text{for } 2 \leq t \leq 6 \end{cases} $$ The graph of this solution for the interval \(0 \leq t \leq 6\) will show an abrupt change in the slope of the solution at \(t=2\) due to the presence of the delta function in the ODE.

Step-by-step solution

Write down the given ODE and initial conditions

The given ODE and initial conditions are as follows: \(y^{\prime \prime}+2y^{\prime}+y= \delta (t-2), \quad y(0) = 0, \quad y^{\prime}(0) =1\)

Assuming the general solution

First, we consider the homogeneous ODE without the delta function on the right side, i.e., \(y^{\prime \prime}+2y^{\prime}+y= 0\) The characteristic polynomial for this homogeneous ODE is: \(r^2 + 2r + 1 = 0\) Solving the quadratic equation, we find that the roots are r=-1 (with multiplicity 2). So the general solution for the homogeneous ODE is given by: \(y_h(t) = C_1 e^{-t}+C_2 te^{-t}\)

Finding the particular solution

Now, we need to find the particular solution of the inhomogeneous ODE using the delta function: \(y^{p\prime\prime} + 2y^{p\prime} + y^p = \delta(t-2)\) Let's assume that the solution is in the following form: $y^p(t) = \begin{cases} Ae^{-t} + Bte^{-t} & \quad \text{for } 0 \leq t < 2\\\\ Ce^{-t} + Dte^{-t} & \quad \text{for } 2 \leq t \leq 6 \end{cases}$

Finding coefficients for the particular solution using continuity and initial jump conditions

Given that Laplace transform of \(\delta(t - 2)\) is \(e^{-2s}\), we compute the Laplace transform of the above assumed solution and equate them to find the coefficients A, B, C, and D using initial conditions and continuity properties. At t=0, we have \(y(0)=0\), which implies: \(A = 0\) So, the particular solution becomes: $y^p(t) = \begin{cases} Bte^{-t} & \quad \text{for } 0 \leq t < 2\\\\ Ce^{-t} + Dte^{-t} & \quad \text{for } 2 \leq t \leq 6 \end{cases}$ Next, the continuity of the function at \(t = 2\) implies: \(2Be^{-2} = 2Ce^{-2}\) This gives us \(B = C\). The initial jump condition for delta function is given by: \(y^{p\prime}(2^+) - y^{p\prime}(2^-) = 1\) Differentiate the above expressions for y with respect to t and compute this equation to obtain the value of D. \([-2e^{-2}(B-D)] - e^{-2}(B-D) = 1\) This gives us \(D = B-1\)

Finding the final particular solution using coefficients

Substitute the values of coefficients obtained in the assumed expression for the particular solution. $y^p(t) = \begin{cases} Bte^{-t} & \quad \text{for } 0 \leq t < 2\\\\ Be^{-t} + (B-1)te^{-t} & \quad \text{for } 2 \leq t \leq 6 \end{cases}$

Write the total solution and apply initial conditions

The total solution is the sum of the homogeneous and the particular solutions. \(y(t) = y_h(t)+y^p(t)\) At t=0, we have \(y^{\prime}(0)=1\). Calculate the derivative \(y^{\prime}(t)\) for the total solution and use this condition to find the value of B. After computing the derivative, we have: \((-B)+(-2Be^{-2}) = 1\) Solving for B results in \(B = -\frac{1}{3}e^2\)

Finalize the total solution

Substitute the value of B obtained in the total solution to get the final expression for y(t). $y(t) = \begin{cases} -\frac{1}{3}e^2te^{-t} & \text{for } 0 \leq t < 2\\\\ -\frac{1}{3}e^2e^{-t} + (\frac{2}{3}e^2 - 1)te^{-t} & \quad \text{for } 2 \leq t \leq 6 \end{cases}$

Graph the obtained solution on the given interval

Finally, the graph of the given solution for the interval \(0 \leq t \leq 6\) can be plotted using a computational tool (e.g., Mathematica, MATLAB, etc.) to visualize the behavior of the solution and how it responds to the delta function input at t=2. The graph will show an abrupt change in the slope of the solution at \(t=2\) due to the presence of the delta function in the ODE.

Key concepts

Delta Function

The delta function, often represented as \(\delta(t-c)\), is a special mathematical function characterized by a spike at \(t = c\). It acts as an impulse or a very short, infinitely high signal with an area of one. Although not a function in the classic sense, it's useful in modeling instantaneous events in physical systems such as a sudden force applied at a point in time.

One property of the delta function is that integrating it over an interval containing \(c\) gives 1, and integrating at areas outside of \(c\) gives 0. This makes it a powerful tool for simplifying complex equations by isolating impacts at specific moments.

• Important in control systems, signal processing, and differential equations.
• Helps in transforming real-world phenomena into mathematically manageable models.

In the context of differential equations, like the one we solved, the delta function represents an external force impacting the system at a single point in time.

Initial Value Problem

An initial value problem (IVP) is a differential equation accompanied by specified values at the start (or another point) of the interval. These starting values are called initial conditions and are crucial for determining a unique solution to the equation.

In this exercise, we have initial conditions at \(t = 0\) given as \(y(0) = 0\) and \(y'(0) = 1\). These conditions allow us to specify the particular solution of the problem by eliminating any ambiguity caused by general solutions.

• These conditions describe the system's state at the beginning of the analysis.
• Often represent physical aspects like initial velocity or displacement in mechanical systems.

Initial value problems often lead to unique solutions for ordinary differential equations (ODEs), ensuring that the solution aligns with the physical or theoretical conditions posed at \(t = 0\).

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms depend on the function and its derivatives alone, without external forces or inputs. This leads to solutions that include constants reflecting the system's natural state. The equation \(y'' + 2y' + y = 0\) is a typical example.

The solution to such equations usually involves finding the roots of the corresponding characteristic equation. If \(r^2 + 2r + 1 = 0\) is the characteristic equation, solving for \(r\) gives roots of -1 with multiplicity 2, leading to a general solution:

• \(y_h(t) = C_1 e^{-t} + C_2 te^{-t}\)

These solutions represent natural decay or growth rates in the system, depending on the nature of the roots, and are critical in forming the basis for the total solution when addressing non-homogeneous problems.

Inhomogeneous Differential Equation

An inhomogeneous differential equation includes an external force or input, such as a delta function, making the solutions more complex than their homogeneous counterparts. The equation given in the problem is \(y'' + 2y' + y = \delta(t-2)\).

To solve this, we find a particular solution that fits the non-homogeneous component. This involves understanding how the input affects the system, often requiring assumptions about the solution and using continuity conditions.

• The Laplace Transform is helpful here, transforming the differential equation into an algebraic equation.
• Differentiate both sides to apply initial jump conditions for accuracy in solutions.

The resultant solution, a combination of the homogeneous and particular solutions, responds to the given initial conditions and the additional external force, providing a complete picture of the system's behavior.