Question

(a) If \(a r^{2}+b r+c=0\) has equal roots \(r_{1},\) show that $$ L\left[e^{r t}\right]=a\left(e^{r t}\right)^{\prime \prime}+b\left(e^{r t}\right)^{\prime}+c e^{r t}=a\left(r-r_{1}\right)^{2} e^{r t} $$ Since the right side of Eq. (i) is zero when \(r=r_{1},\) it follows that \(\exp \left(r_{1} t\right)\) is a solution of \(L[y]=a y^{\prime \prime}+b y^{\prime}+c y=0\) (b) Differentiate Eq. (i) with respect to \(r\) and interchange differentiation with respect to \(r\) and with respect to \(t\), thus showing that $$ \frac{\partial}{\partial r} L\left[e^{r t}\right]=L\left[\frac{\partial}{\partial r} e^{r t}\right]=L\left[t e^{r t}\right]=a t e^{r t}\left(r-r_{1}\right)^{2}+2 a e^{r t}\left(r-r_{1}\right) $$ Since the right side of \(\mathrm{Eq}\). (ii) is zero when \(r=r_{1},\) conclude that \(t \exp \left(r_{1} t\right)\) is also a solution of \(L[y]=0 .\)

Short answer

Question: Show that if the quadratic equation \(ar^2 + br + c = 0\) has equal roots \(r_1\), then (a) Solve the ODE \(L\left[e^{rt}\right]=a(r-r_1)^2 e^{rt}\), (b) Show that \(L\left[t\exp(r_1t)\right]=0\). Answer: (a) We have proved that if the quadratic equation \(ar^2 + br + c = 0\) has equal roots \(r_1\), then the given equation \(L\left[e^{rt}\right]=a(r-r_1)^2 e^{rt}\) holds. (b) We have shown that when \(r=r_1\), the function \(t\exp(r_1t)\) satisfies the given ODE, so \( L\left[t\exp(r_1t)\right]=0\).

Step-by-step solution

Find the derivatives of the exponential function

First, let's find the first and second derivatives of the function \(e^{rt}\) with respect to \(t\). We get: 1. First derivative: $$ \frac{d}{dt}(e^{rt}) = r e^{rt} $$ 2. Second derivative: $$ \frac{d^2}{dt^2}(e^{rt}) = r^2 e^{rt} $$ Now, we will use these derivatives in the given equation.

Plug the derivatives into the equation

Substitute the first and second derivatives found in step 1 into the left-hand side of the equation: $$ L\left[e^{rt}\right] = a\left(e^{rt}\right)^{\prime\prime} + b\left(e^{rt}\right)^{\prime} + ce^{rt} = ar^2 e^{rt} + bre^{rt} + ce^{rt} $$ We are given that the quadratic equation \(ar^2 + br + c = 0\) has equal roots \(r_1\), so we can substitute \(r=r_1\) in the above equation. Thus, we have: $$ a\left(r-r_1\right)^2 e^{rt} = 0 $$ Since the left side matches the given right side of the equation in part (a), we have proved the required result.

Differentiate with respect to \(r\)

Now, let's differentiate both sides of the equation \(L[e^{rt}]\) with respect to \(r\): $$ \frac{\partial}{\partial r} L\left[e^{rt}\right] = \frac{\partial}{\partial r}(ar^2 e^{rt} + bre^{rt} + ce^{rt}) $$

Interchange the order of differentiation

Next step is to interchange the order of differentiation. For this, let's recall the second derivative found in step 1: To find \(\frac{\partial}{\partial r} L[e^{rt}]\), we can first rewrite it as: \(L[\frac{\partial}{\partial r} e^{rt}]\). Now we can plug in the first and second derivatives of \(e^{rt}\) and differentiate them with respect to \(r\), while also considering the initial expression we got in step 2: $$ L\left[\frac{\partial}{\partial r} e^{rt}\right] = L\left[t e^{rt}\right] = a t e^{rt}\left(r-r_1\right)^ 2 + 2 a e^{rt}\left(r-r_1\right) $$ We are given that the equation \(ar^2+br+c=0\) has equal roots \(r_1\). As a result, when we substitute \(r=r_1\) in the expression we derived in step 4, the right side becomes zero. Therefore, when \(r=r_1\), the function \(t\exp(r_1t)\) also satisfies the given ODE, fulfilling the desired conclusion for part (b).

Key concepts

Characteristic Equation

The characteristic equation plays a crucial role in solving linear ordinary differential equations (ODEs), especially homogeneous ODEs. It is a polynomial equation obtained by replacing the derivative of the function with a polynomial in terms of some variable, usually denoted by 'r'. In the context of our problem, the characteristic equation is given by \(a r^{2} + b r + c = 0\). When this equation has equal roots, denoted \(r_1\), it indicates that the differential equation has a solution that includes terms involving the exponential function raised to \(r_1t\).

Understanding the implications of the characteristic equation provides insight into the nature of the ODE's solutions. In this case, equal roots suggest repetitive roots in the solution, which leads to a particular structure of the solution set, often involving multiple exponential terms and their derivatives.

Exponential Functions

Exponential functions are a type of mathematical function of the form \(e^{rt}\), where 'e' is the base of the natural logarithm, 'r' is a constant, and 't' is the variable. These functions are pivotal in solving differential equations because they exhibit the property \( \frac{d}{dt}{e^{rt}} = re^{rt} \), which means their derivative is proportional to the function itself.

When solving a homogeneous ODE with constant coefficients, the exponential function is a powerful tool to construct the complementary function. This is evident in the given exercise, where the derivative properties of exponential functions are utilized to verify the solutions of the differential equation.

Ordinary Differential Equations (ODEs)

Ordinary differential equations are equations that relate functions with their derivatives. They are termed 'ordinary' to distinguish them from partial differential equations, which involve partial derivatives with respect to multiple variables. ODEs are classified based on order, linearity, and homogeneity. The simplest form of an ODE is a first-order equation, but they can be of any order.

In the exercise we're examining, the focus is on second-order linear homogeneous ODEs, characterized by an equation of the form \(ay'' + by' + cy = 0\), where \(y'\) and \(y''\) are the first and second derivatives of the function \(y(t)\). The solutions to these equations describe a variety of physical phenomena, ranging from motion to heat transfer.

Homogeneous Equations

A homogeneous equation in the realm of differential equations is one where every term involves the function or its derivatives, and there is no 'free' term (i.e., a term without the function or its derivatives). Homogeneous ODEs, such as the one in our exercise \(ay'' + by' + cy = 0\), are characterized by the superposition principle, meaning that any linear combination of solutions is also a solution.

Solving Homogeneous Equations

To solve a homogeneous ODE, one typically assumes a solution of the form \(e^{rt}\), substitutes it into the equation, and determines the values of 'r' via the characteristic equation. The nature of the roots (real, repeated, or complex) influences the structure of the general solution.

Derivative

The derivative of a function represents the rate at which the function is changing at any given point. It's a fundamental concept in calculus that is extensively used in various domains of mathematics and science. For exponential functions, such as \(e^{rt}\), the derivative provides a way to express growth or decay dynamics, which is especially useful in the context of differential equations.

In the exercise, differentiation with respect to the parameter 'r' is used to derive an additional solution for the ODE. By differentiating the assumed solution \(e^{rt}\) with respect to 'r', we acquire a broader set of solutions, which include functions like \(te^{rt}\), when the initial characteristic equation has repeated roots.