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}=8(t-1) \quad y(0)=1 \quad y^{\prime}(0)=0 \quad 0

Short answer

#Question# Given the initial value problem: \(y'' - 2y' = 8(t-1)\) with initial conditions \(y(0) = 1\) and \(y'(0) = 0\), determine the particular solution that satisfies these conditions and graph the solution on the interval \(0 < t < 2\). #Answer# The particular solution that satisfies the initial conditions is \(y(t) = 1 - e^{2t} - 2t^2 + 2t\). Graph the solution on the interval \(0 < t < 2\) using a graphing tool to visualize the behavior of the solution over time.

Step-by-step solution

Solve the Homogeneous Equation

To solve the homogeneous equation associated with the given ODE, we need to find the solutions to $$y^{\prime \prime} - 2y^{\prime} = 0$$ This is a second-order linear ODE with constant coefficients. We assume a solution in the form of \(y=e^{rt}\) and substitute it into the homogeneous equation: $$r^2 e^{rt} - 2re^{rt} = 0$$ Factor out \(e^{rt}\) and solve the remaining quadratic equation: $$r(r-2)=0$$ This has two roots: \(r = 0\) and \(r = 2\). Therefore, the homogeneous solution is: $$y_h = C_1 + C_2 e^{2t}$$

Determine a Particular Solution

Now we need to find a particular solution to the given ODE: $$y^{\prime \prime} - 2y^{\prime} = 8(t-1)$$ We want to find a specific solution in the form \(y_p = At^2 + Bt + C\). By taking the derivatives, we obtain: $$y_p^{\prime} = 2At + B$$ $$y_p^{\prime \prime} = 2A$$ Now, we can substitute these expressions into the ODE: $$2A - 2(2At + B) = 8(t-1)$$ Simplifying: $$2A - 4At - 2B = 8t - 8$$ Equating coefficients: $$\begin{cases} -4A = 8 \\ -2B + 2A = -8 \end{cases}$$ Solving, we get \(A=-2\) and \(B=2\). Thus, the particular solution is: $$y_p = -2t^2 + 2t$$

Add the Homogeneous and Particular Solutions

We now combine the homogeneous and particular solutions to find the general solution of the ODE: $$y(t) = C_1 + C_2 e^{2t} - 2t^2 + 2t$$

Apply the Initial Conditions

Given the initial conditions \(y(0) = 1\) and \(y^{\prime}(0) = 0\), we can find the particular solution that satisfies them. For \(y(0) = 1\): $$y(0) = C_1 + C_2 - 0 + 0 = 1$$ For \(y^{\prime}(0) = 0\), notice that \(y^{\prime}(t) = -4t + 2 + 2C_2e^{2t}\). Thus, we have: $$y^{\prime}(0) = -0 + 2 + 2C_2 = 0$$ From these two equations, we get \(C_1 = 1\) and \(C_2 = -1\). Therefore, the particular solution that satisfies the initial conditions is: $$y(t) = 1 - e^{2t} - 2t^2 + 2t$$

Graph the Solution on the Indicated Interval

To graph the solution, \(y(t) = 1 - e^{2t} - 2t^2 + 2t\) on the interval \(0 < t < 2\), we would use a graphing tool (such as a graphing calculator or online software) to plot the function within the specified range. The graph will show the behavior of the solution over time, satisfying the initial conditions, and providing insight into the system's response to the given input.

Key concepts

Differential Equations

Differential equations are mathematical tools that describe the relationship between a function and its derivatives. In the context of physical, engineering, and biological systems, differential equations are used to model dynamic processes where change is constant.

An example of a differential equation is the one given in our exercise: \(y'' - 2y' = 8(t-1)\). This second-order linear differential equation indicates that the acceleration of the system (the second derivative of \(y\)) minus twice the velocity of the system (the first derivative of \(y\)) equals a forcing function that depends on time \(t\). The initial conditions \(y(0)=1\) and \(y'(0)=0\) provide specific values for the function and its first derivative at the starting time \(t=0\), allowing us to tailor the solution to a particular scenario.

Delta Functions

Delta functions, especially the Dirac delta function, are a fundamental concept in engineering and physics, acting as idealized representations of impulses or very sharp spikes over infinitesimally short intervals. These functions have the unique property that they are zero everywhere except at the point of impulse, where they are infinite, and the integral over the entire function's domain equals one.

In the context of the exercise, the given differential equation does not directly involve a delta function. However, the implication of the delta function can be seen in real-world applications of such equations, where inputs of large amplitude and short duration might be idealized as delta functions. These act as sudden forces or inputs to the system that the differential equation is modeling.

Homogeneous Equations

Homogeneous equations are a subset of differential equations where all the terms are a function of the dependent variable and its derivatives. A homogeneous equation can be written in the form \(f(y, y', y'', ..., y^{(n)}) = 0\), where each term includes the dependent variable \(y\) or one of its derivatives.

The homogeneous part of an equation often corresponds to the system's natural behavior without external forces. In our case, the homogeneous equation \(y'' - 2y' = 0\) describes the system's response in the absence of the external forcing term \(8(t-1)\). Solving the homogeneous equation provides us with the complementary (or homogeneous) solution, which we denote as \(y_h\). This solution lays the foundation upon which the particular solution, representing the response to the external force, is added.

Particular Solution

A particular solution (\(y_p\)) to a differential equation is one that satisfies not only the differential equation itself but also specific initial or boundary conditions. Unlike the general solution, which includes arbitrary constants to encapsulate a family of solutions, the particular solution is unique to the given conditions.

In the step-by-step exercise, finding a particular solution requires making an educated guess about the form of the solution—this is often termed as the method of undetermined coefficients. In our specific example, \(y_p\) was assumed to be a quadratic polynomial \(At^2 + Bt + C\) due to the nature of the non-homogeneous term \(8(t-1)\). After derivation and substitution into the original differential equation, the coefficients \(A\), \(B\), and \(C\) are determined to satisfy the equation. This particular solution reflects the system's predictable response to a known force or input.