Question

\begin{array}{l} \text { Determine the solution of each of the following initial-value problems: }\\\ \text { (a) } \quad y^{\prime \prime}+4 y=\cos \omega t-\delta(t-1) \text { , }\\\ \begin{array}{l} y(0)=0, \quad y^{\prime}(0)=0 . \\ \text { (b) } \quad y^{\prime \prime \prime}+y^{\prime}=8(t-2)+e^{-t}, \end{array}\\\ y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0 . \end{array}

Short answer

Answer: For the second-order ODE, the homogeneous solution is \(y_h(t) = c_1 \sin(2t) + c_2 \cos(2t)\) and the particular solution has the form \(y_p(t) = y_p^1(t) + y_p^2(t)\), where \(y_p^1(t) = A \cos(\omega t) + B \sin(\omega t)\) and \(y_p^2(t)\) is found using the Laplace transform or Green's function approach. For the third-order ODE, the homogeneous solution is given by \(y_h(t) = c_1 + c_2 \sin(\sqrt{3}t) + c_3 \cos(\sqrt{3}t)\) and the particular solution has the form \(y_p(t) = y_p^1(t) + y_p^2(t)\), where \(y_p^1(t) = At + B\) and \(y_p^2(t) = Ce^{-t}\).

Step-by-step solution

a) Second-order ODE

: We will solve the second-order ODE \(y'' + 4y = \cos(\omega t) - \delta (t - 1)\) subject to the initial conditions \(y(0) = 0\) and \(y'(0) = 0\). Step 1: Solve the homogeneous equation: First, we need to solve the homogeneous equation \(y'' + 4y = 0\). This has the characteristic equation \(r^2 + 4 = 0\) which has solutions \(r = \pm 2i\). So the homogeneous solution is given by \(y_h(t) = c_1 \sin(2t) + c_2 \cos(2t)\). Step 2: Find the particular solution: Now we need to find the particular solution \(y_p(t)\) for the non-homogeneous equation. For \(\cos(\omega t)\) term, we try the solution form \(y_p^1(t) = A \cos(\omega t) + B \sin(\omega t)\) and solve for A and B. For the \(\delta(t-1)\) term, we have a Laplace transform solution or Green's function approach as detailed in textbooks. Let's denote this solution as \(y_p^2(t)\). The overall particular solution is the sum of the two: \(y_p(t) = y_p^1(t) + y_p^2(t)\). Step 3: Determine the constants: Use the initial conditions \(y(0) = 0\) and \(y'(0) = 0\) to find the constants \(c_1\) and \(c_2\). Finally, the solution to the given ODE is \(y(t) = y_h(t) + y_p(t)\).

b) Third-order ODE

: We will solve the third-order ODE \(y''' + y' = 8(t - 2) + e^{-t}\) subject to the initial conditions \(y(0) = 1\), \(y'(0) = 0\), and \(y''(0) = 0\). Step 1: Solve the homogeneous equation: First, we need to solve the homogeneous equation \(y''' + y' = 0\). This has the characteristic equation \(r^3 + r = 0\). It has roots \(r = 0\), \(r = \sqrt{3}i\), and \(r = -\sqrt{3}i\). So the homogeneous solution is given by \(y_h(t) = c_1 + c_2 \sin(\sqrt{3}t) + c_3 \cos(\sqrt{3}t)\). Step 2: Find the particular solution: Now we need to find the particular solution \(y_p(t)\) for the non-homogeneous equation. We will use the method of undetermined coefficients. For the \(8(t - 2)\) term, we try the solution form \(y_p^1(t) = At + B\) and solve for A and B. For the \(e^{-t}\) term, we try the solution form \(y_p^2(t) = Ce^{-t}\) and solve for C. The overall particular solution is the sum of the two: \(y_p(t) = y_p^1(t) + y_p^2(t)\). Step 3: Determine the constants: Use the initial conditions \(y(0) = 1\), \(y'(0) = 0\), and \(y''(0) = 0\) to find the constants \(c_1\), \(c_2\), and \(c_3\). Finally, the solution to the given ODE is \(y(t) = y_h(t) + y_p(t)\).

Key concepts

Initial-Value Problems

An initial-value problem (IVP) in the context of ordinary differential equations is a problem where not only an equation is given, but also the values of the unknown function and its derivatives at a specific point, typically at time t=0. These initial conditions are crucial as they allow us to determine the constants that arise in the solutions of differential equations, which helps in finding the unique solution that fits the specific problem.

For example, in the given exercise, for part a), the IVP is defined by the differential equation \(y'' + 4y = \cos(\omega t) - \delta(t - 1)\) along with initial conditions \(y(0) = 0\) and \(y'(0) = 0\). These conditions provide us with enough information to determine the specific constants for the homogeneous solution upon solving the associated characteristic equation.

Characteristic Equation

The characteristic equation is a tool we use to solve linear homogeneous differential equations. It is obtained by substituting a trial solution of the form \( e^{rt} \) into the homogeneous part of the differential equation, resulting in a polynomial where r represents the roots. These roots are essential as they help to construct the general solution.

In part a) of our problem, the corresponding characteristic equation is \(r^2 + 4 = 0\), which yields complex roots \(\pm 2i\). These roots indicate a solution composed of sinusoidal functions, which you will find in the homogeneous solution.

Homogeneous Solutions

Homogeneous solutions correspond to solving the differential equation assuming the absence of non-homogeneous parts (i.e., when the equation equals zero). They are crucial to the overall solution because they represent the system's natural behavior, devoid of external influences.

In our exercise, we found that the homogeneous solution of part a) is given by \(y_h(t) = c_1 \sin(2t) + c_2 \cos(2t)\) by solving \(y'' + 4y = 0\), using the roots from the characteristic equation.

Particular Solutions

While particular solutions are obtained by finding a function that satisfies the original non-homogeneous differential equation. They are not general solutions because they do not include arbitrary constants; rather, they address the specific form of the non-homogeneous term. It's like finding a puzzle piece that precisely fits one slot in the puzzle.

In the provided steps, particular solutions are found by assuming a trial solution based on the form of the non-homogeneous part, such as a polynomial, exponential, or trigonometric function, and then determining the coefficients that make it a true solution.

Method of Undetermined Coefficients

The method of undetermined coefficients is an approach for finding particular solutions to non-homogeneous linear differential equations with constant coefficients. The method involves guessing a form for the particular solution based on the non-homogeneous term and plugging it back into the equation to determine the unknown coefficients.

In part b) of our exercise, the suspected forms for the solution were \(y_p^1(t) = At + B\) to address the \(8(t - 2)\) term and \(y_p^2(t) = Ce^{-t}\) for the \(e^{-t}\) term. Algebraic work then provided the particular solutions once we plugged in and equated coefficients.

Laplace Transform

The Laplace transform is a powerful integral transformation technique used to convert differential equations into algebraic equations, which are often easier to solve. It particularly shines when handling initial-value problems with discontinuous or impulsive functions such as the Dirac delta function mentioned in our exercise.

The transform pairs a function in the time domain, t, with a function in the complex frequency domain, typically denoted as s. After applying the Laplace Transform to an IVP, one solves the equation for the transformed variable and then applies the inverse transform to get back to the time-domain solution.