Question

Follow the instructions in Problem 28 to solve the differential equation $$ y^{\prime \prime}+2 y^{\prime}+5 y=\left\\{\begin{array}{ll}{1,} & {0 \leq t \leq \pi / 2} \\ {0,} & {t>\pi / 2}\end{array}\right. $$ $$ \text { with the initial conditions } y(0)=0 \text { and } y^{\prime}(0)=0 $$ $$ \begin{array}{l}{\text { Behavior of Solutions as } t \rightarrow \infty \text { , In Problems } 30 \text { and } 31 \text { we continue the discussion started }} \\ {\text { with Problems } 38 \text { through } 40 \text { of Section } 3.5 \text { . Consider the differential equation }}\end{array} $$ $$ a y^{\prime \prime}+b y^{\prime}+c y=g(t) $$ $$ \text { where } a, b, \text { and } c \text { are positive. } $$

Short answer

Question: Find the specific solution for the given second-order linear differential equation with initial conditions: $$ y^{\prime \prime} + 2 y^{\prime} + 5 y = g(t) $$ where $$ g(t) = \begin{cases} 1, & \text{if } 0 \leq t \leq \pi/2, \\ 0, & \text{if } t> \pi/2. \end{cases} $$ and $$ y(0) = 0, \quad y^{\prime}(0) = 0. $$ Answer: The specific solution for the given second-order linear differential equation with initial conditions is: $$ y(t) = \begin{cases} \frac{1}{5}, & 0 \leq t \leq \pi/2, \\ 0, & t > \pi/2. \end{cases} $$

Step-by-step solution

Analysis of the given differential equation

The given differential equation is: $$ y^{\prime \prime} + 2 y^{\prime} + 5 y = g(t) $$ where $$ g(t) = \begin{cases} 1, & \text{if } 0 \leq t \leq \pi/2, \\ 0, & \text{if } t> \pi/2. \end{cases} $$ and the initial conditions are $$ y(0) = 0, \quad y^{\prime}(0) = 0. $$

Solve the homogeneous equation

First, we look for the general solution of the homogeneous equation associated with the given non-homogeneous equation: $$ y^{\prime \prime} + 2 y^{\prime} + 5 y = 0 $$ This is a second-order linear homogeneous equation with constant coefficients. We can find the characteristic equation associated with this homogeneous equation: $$ r^2 + 2r + 5 = 0 $$ We need to find the roots of this characteristic equation. Using the quadratic formula, we get: $$ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 5}}{2} = -1 \pm 2i $$ So, the roots of the characteristic equation are complex conjugates: \(r_1 = -1 + 2i\) and \(r_2 = -1 - 2i\). The general solution of the homogeneous equation is given by: $$ y_h(t) = c_1 e^{(-1 + 2i)t} + c_2 e^{(-1 - 2i)t} $$

Determine the particular solution of non-homogeneous equation

We break down the problem into two cases: when \(0 \leq t \leq \pi / 2\) and when \(t > \pi / 2\). Case 1: \(0 \leq t \leq \pi / 2\) where \(g(t) = 1\) We guess a constant particular solution \(y_p(t) = A\) for the non-homogeneous equation. Taking the derivatives, we have \(y_p^{\prime}(t) = 0\) and \(y_p^{\prime \prime}(t) = 0\). Substituting these values back into the non-homogeneous equation, we get: $$ 0+2(0)+5(A)=1 $$ Solving for A, we get \(A = \frac{1}{5}\). So, the particular solution for this case is given by: $$ y_p(t) = \frac{1}{5} $$ Case 2: \(t>\pi/2\) where \(g(t) = 0\) For this case, the particular solution is simply 0. Now, the overall particular solution for the given non-homogeneous equation is: $$ y_p(t) = \begin{cases} \frac{1}{5}, & 0 \leq t \leq \pi/2, \\ 0, & t > \pi/2. \end{cases} $$

Combine the general solutions

The general solution of the given non-homogeneous equation can be found by adding the homogeneous and particular solutions: $$ y(t) = y_h(t) + y_p(t) = c_1 e^{(-1 + 2i)t} + c_2 e^{(-1 - 2i)t} + y_p(t) $$

Apply the initial conditions

We have the initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 0\). Let us apply these conditions to find the specific solution. Applying the first initial condition \(y(0) = 0\): $$ y(0) = c_1 e^{(0)} + c_2 e^{(0)} + 0 = 0 \Rightarrow c_1 + c_2 = 0 \Rightarrow c_2 = -c_1 $$ Now, let us find the first derivative of \(y(t)\): $$ y^{\prime}(t) = (-1 + 2i)c_1 e^{(-1 + 2i)t} + (-1 - 2i)c_2 e^{(-1 - 2i)t} + y_p^{\prime}(t) $$ Applying the second initial condition \(y^{\prime}(0) = 0\): $$ y^{\prime}(0) = (-1 + 2i)c_1 e^{(0)} + (-1 - 2i)c_2 e^{(0)} + 0 = 0 $$ Substituting \(c_2=-c_1\), we get: $$ (-1 + 2i)c_1 + (-1 - 2i)(-c_1) = 0 $$ Solving for \(c_1\), we find that \(c_1 = 0\). Therefore, \(c_2 = 0\) as well.

Determine the specific solution

Now that we have found the values of \(c_1\) and \(c_2\), we can write the specific solution for the given non-homogeneous equation: $$ y(t) = 0\cdot e^{(-1 + 2i)t} + 0\cdot e^{(-1 - 2i)t} + y_p(t) = y_p(t) $$ So, the specific solution for the given non-homogeneous equation is: $$ y(t) = \begin{cases} \frac{1}{5}, & 0 \leq t \leq \pi/2, \\ 0, & t > \pi/2. \end{cases} $$

Key concepts

Characteristic Equation

The characteristic equation plays a crucial role in solving linear homogeneous differential equations with constant coefficients. It is a polynomial whose roots are used to construct the general solution of the differential equation. For example, given a second-order differential equation like
\( y^{\prime \prime} + p y^{\prime} + q y = 0 \),
the characteristic equation takes the form
\( r^2 + pr + q = 0 \).
The solutions to this polynomial, known as the characteristic roots, determine the behavior of the differential equation's solutions.

Complex Roots

When the roots are complex, as in our exercise step, they appear in conjugate pairs, say \( r_1 = a + bi \) and \( r_2 = a - bi \), leading to solutions involving exponential functions multiplied by sine and cosine functions, reflecting oscillatory behavior. In the exercise, the characteristic equation was found to be
\( r^2 + 2r + 5 = 0 \),
which has the roots \( -1 + 2i \) and \( -1 - 2i \), indicating an oscillatory solution with a damping factor due to the real part of the roots.

Homogeneous Solutions

Homogeneous solutions, denoted as \( y_h(t) \), address the part of the differential equation without external forcing, meaning the right-hand side is zero. These solutions are derived purely from the differential equation's coefficients and its initial conditions but are independent of any external influences described by a function like \( g(t) \).

Handling Complex Conjugate Roots

The presence of complex conjugate roots in the characteristic equation leads us to a solution involving complex exponentials. However, we usually represent these solutions in terms of sine and cosine functions because these are real functions, and we often deal with real-valued problems. With initial conditions, these solutions will constitute a family of curves, where each member of the family satisfies the homogeneous differential equation.

Particular Solutions

Particular solutions, \( y_p(t) \), are found by considering the non-homogeneous part of the differential equation, that is, the external forcing or input represented by \( g(t) \). Unlike the homogeneous solution, which depicts the system's natural behavior, the particular solution captures the response specifically due to \( g(t) \).

Finding Particular Solutions

The method of undetermined coefficients is often utilized to guess a form for \( y_p(t) \) based on the type of function that \( g(t) \) is. In our exercise, \( g(t) \) is a piecewise function, changing its form at \( t = \pi / 2 \), which led to different particular solutions for the intervals \( 0 \leq t \leq \pi / 2 \) and \( t > \pi / 2 \). The particular solution reflects only the influence of this external force and ignores the system's natural dynamics.

Initial Conditions

Initial conditions are the starting values given at a certain time, typically at \( t=0 \), for a differential equation. They provide a snapshot of the system's state at a specific instant and are crucial for determining the particular solution to a differential equation.

Applying Initial Conditions

In our exercise, we were given \( y(0) = 0 \) and \( y^{\text{'}}(0) = 0 \). These conditions are used to find the unique coefficients \( c_1 \) and \( c_2 \) of the homogeneous solution. Essentially, they anchor the solution to a specific curve that satisfies both the differential equation and the initial conditions at \( t = 0 \). Without initial conditions, we would have infinitely many solutions to a differential equation, corresponding to a continuum of possible states of the system at the start of observation.