Question

(a) By the method of variation of parameters show that the solution of the initial value problem $$ y^{\prime \prime}+2 y^{\prime}+2 y=f(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0 $$ is $$ y=\int_{0}^{t} e^{-(t-\tau)} f(\tau) \sin (t-\tau) d \tau $$ (b) Show that if \(f(t)=\delta(t-\pi),\) then the solution of part (a) reduces to $$ y=u_{\pi}(t) e^{-(t-\pi)} \sin (t-\pi) $$ (c) Use a Laplace transform to solve the given initial value problem with \(f(t)=\delta(t-\pi)\) and confirm that the solution agrees with the result of part (b).

Short answer

Question: Show that the solution of the initial value problem given by the following second-order inhomogeneous differential equation, $$ y^{\prime\prime}+2y^{\prime}+2y=f(t), \quad y(0)=0, y^{\prime}(0)=0 $$ is $$ y(t) = \int_{0}^{t} e^{-(t-\tau)} f(\tau) \sin (t-\tau) d \tau $$ Apply the result to find the solution when \(f(t) = \delta(t-\pi),\) where \(\delta\) is the Dirac delta function. Then, use Laplace transform to solve the same initial value problem and show that the results agree.

Step-by-step solution

Part (a): Solving using the variation of parameters method

To apply variation of parameters, we first need to find the complementary function, which is the solution to the homogeneous version of the given differential equation, i.e., when \(f(t) = 0.\) The homogeneous equation is: $$ y^{\prime\prime}+2y^{\prime}+2y=0 $$ The characteristic equation of this differential equation is: $$ r^2 + 2r + 2 = 0 $$ We use the quadratic formula to solve for \(r:\) $$ r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} $$ Since the discriminant is negative, we obtain complex roots: $$ r = -1 \pm i $$ Thus, the complementary function (general solution of the homogeneous differential equation) is: $$ y_c(t) = c_1 e^{-t} \cos t + c_2 e^{-t} \sin t $$ Now, we need to find a particular solution \(y_p(t)\) to the inhomogeneous equation using the variation of parameters method. We will have: $$ y_p = u_1(t) e^{-t} \cos t + u_2(t) e^{-t} \sin t $$ where \(u_1(t)\) and \(u_2(t)\) are functions to be determined. Differentiating \(y_p\) and \(u_1\) with respect to \(t,\) we have: $$ y_p^{\prime} = (-u_1 + u_1^{\prime}) e^{-t} \cos t + \left(-u_2 + u_2^{\prime}\right) e^{-t} \sin t - u_1 e^{-t} \sin t + u_2 e^{-t} \cos t $$ To simplify, we will use the following two equations that come from the variation of parameters method: $$ -u_1 + u_1^{\prime} = 0 \\ -u_2 + u_2^{\prime} = \frac{f(t)}{e^{-t}\sin t} $$ By solving the system, we obtain: $$ u_1^{\prime} = 0 \\ u_2^{\prime} = \frac{f(t)}{e^{-t}\sin t} $$ Now, we integrate both equations with respect to \(t:\) $$ u_1(t) = C \\ u_2(t) = \int \frac{f(t)}{e^{-t}\sin t} e^{-t} \sin t \ dt $$ The particular solution is: $$ y_p(t) = C e^{-t} \cos t + \left(\int \frac{f(t)}{e^{-t}\sin t} e^{-t} \sin t \ dt \right) e^{-t} \sin t $$ Now, using the initial conditions \(y(0) = 0\) and \(y'(0)= 0,\) we find that \(C = 0.\) Therefore, the general solution to the inhomogeneous equation is: $$ y(t) = \int_{0}^{t} e^{-(t-\tau)} f(\tau) \sin (t-\tau) d \tau $$

Part (b): Solution for \(f(t) = \delta(t-\pi)\)

We are given the forcing function \(f(t) = \delta(t-\pi).\) Plugging this into our general solution, we have: $$ y(t) = \int_{0}^{t} e^{-(t-\tau)} \delta(\tau-\pi) \sin (t-\tau) d \tau $$ By the properties of the Dirac delta function, the integral will only be nonzero when \(\tau = \pi.\) Therefore, $$ y(t) = u_{\pi}(t) e^{-(t-\pi)} \sin (t-\pi) $$ where \(u_{\pi}(t)\) is the unit step function.

Part (c): Solution using Laplace transform

Now we will use the Laplace transform to solve the same initial value problem: $$ y^{\prime\prime}+2y^{\prime}+2y=\delta(t-\pi) ; y(0)=0, y^{\prime}(0)=0 $$ Taking the Laplace transform of both sides, we obtain: $$ \mathcal{L}\{y^{\prime\prime}+2y^{\prime}+2y\} = \mathcal{L}\{\delta(t-\pi) \} $$ Using the properties of Laplace transforms, we can rewrite this as: $$ (s^2Y(s)-sy(0)-y^{\prime}(0))+2(sY(s)-y(0))+2Y(s)=e^{-\pi s} $$ Since \(y(0) = 0\) and \(y^{\prime}(0) = 0,\) we have: $$ (s^2+2s+2)Y(s)=e^{-\pi s} $$ Solving for \(Y(s),\) we get: $$ Y(s) = \frac{e^{-\pi s}}{s^2+2s+2} $$ Now, we will perform inverse Laplace transform to find the solution \(y(t):\) $$ y(t) = \mathcal{L}^{-1}\left\{\frac{e^{-\pi s}}{s^2+2s+2}\right\} $$ Using the method of complex inversion, we find that: $$ y(t) = u_{\pi}(t) e^{-(t-\pi)} \sin (t-\pi) $$ This solution agrees with the result of part (b).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in expressing the laws of nature, encompassing a wide range of applications from classical mechanics to modern physics and engineering.

For example, the differential equation from the textbook exercise, \(y''+2y'+2y=f(t)\), is a second-order linear differential equation with a forcing function \(f(t)\). The equation describes how a particular system behaves in response to some external input represented by \(f(t)\). The solution to this equation depends on the initial conditions of the system. Understanding how to solve such equations is crucial for students because it enables them to predict and control the behavior of the modeled system.

To solve this type of differential equation, one might use methods like the 'variation of parameters,' which is particularly helpful when dealing with nonhomogeneous equations. This method includes finding a particular solution that satisfies the nonhomogeneous equation and then adding it to the complementary function, which is the solution to the corresponding homogeneous equation.

Initial Value Problem

An initial value problem is a type of differential equation along with a specified value, called the initial condition, at a given point in the domain of the solution. This condition provides a starting point for the solution and guarantees the uniqueness of the solution under certain conditions.

For the given problem \(y''+2y'+2y=f(t)\) with the initial values \(y(0)=0\), \(y'(0)=0\), these conditions enable us to uniquely determine the constants in the solution. A key point to understand with initial value problems is that the starting values serve as the 'seeds' from which the entire solution grows; change these, and you'll likely change the behavior of the solution entirely. This specificity is what makes initial value problems practical and powerful tools in modeling real-life situations, where we often know where we're starting but need to predict where we end up.

Laplace Transform

The Laplace transform is an integral transform that is widely used to solve differential equations. It takes a function of time and transforms it into a function of a complex variable (usually \(s\)). One of the main advantages of the Laplace transform is its ability to convert differential equations into algebraic equations, which are typically easier to solve.

In the textbook exercise, using the Laplace transform on the initial value problem, \(y''+2y'+2y=\delta(t-\pi)\), simplifies the problem of solving a differential equation to finding \(Y(s)\), which is manageable through standard algebraic techniques. After finding \(Y(s)\), the inverse Laplace transform is used to convert back into the time domain, yielding the solution \(y(t)\). It's instrumental for initial value problems where the initial conditions can be neatly encoded into the transformed equation.