Question

Suppose the characteristic equation of $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ has a repeated real root \(r_{1}\). Temporarily, think of \(e^{r x}\) as a function of two real variables \(x\) and \(r\). (a) Show that $$ a \frac{\partial^{2}}{\partial^{2} x}\left(e^{r x}\right)+b \frac{\partial}{\partial x}\left(e^{r x}\right)+c e^{r x}=a\left(r-r_{1}\right)^{2} e^{r x} $$ (b) Differentiate (B) with respect to \(r\) to obtain $$ a \frac{\partial}{\partial r}\left(\frac{\partial^{2}}{\partial^{2} x}\left(e^{r x}\right)\right)+b \frac{\partial}{\partial r}\left(\frac{\partial}{\partial x}\left(e^{r x}\right)\right)+c\left(x e^{r x}\right)=\left[2+\left(r-r_{1}\right) x\right] a\left(r-r_{1}\right) e^{r x} $$ (c) Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain $$ a \frac{\partial^{2}}{\partial x^{2}}\left(x e^{r x}\right)+b \frac{\partial}{\partial x}\left(x e^{r x}\right)+c\left(x e^{r x}\right)=\left[2+\left(r-r_{1}\right) x\right] a\left(r-r_{1}\right) e^{r x} $$ (d) Set \(r=r_{1}\) in (B) and (D) to see that \(y_{1}=e^{r_{1} x}\) and \(y_{2}=x e^{r_{1} x}\) are solutions of (A)

Short answer

Question: Show that a second-order linear homogeneous differential equation with constant coefficients has solutions y1 and y2 when r is equal to r1. Solution: We first plug the exponential function \(e^{rx}\) into the given second-order linear homogeneous differential equation and find a quadratic equation in terms of r. Then, we differentiate the equation with respect to r and reverse the order of the partial derivatives to obtain a new equation. By setting r equal to r1, we find the two solutions y1 and y2 as \(y_{1}=e^{r_{1} x}\) and \(y_{2}=x e^{r_{1} x}\), respectively.

Step-by-step solution

Part (a)

Plug the function \(e^{rx}\) into the given differential equation: $$ a \frac{\partial^{2}}{\partial x^{2}}\left(e^{r x}\right)+b\frac{\partial}{\partial x}\left(e^{r x}\right)+c e^{r x}=0 $$ First, compute the partial derivatives \(\frac{\partial}{\partial x}\left(e^{r x}\right)\) and \(\frac{\partial^{2}}{\partial x^{2}}\left(e^{r x}\right)\): $$ \frac{\partial}{\partial x}\left(e^{r x}\right) = re^{r x} $$ $$ \frac{\partial^{2}}{\partial x^{2}}\left(e^{r x}\right) = r^{2}e^{r x} $$ Substitute these partial derivatives into the equation: $$ a (r^{2}e^{r x})+b (re^{r x})+c e^{r x} = a\left(r-r_{1}\right)^{2} e^{r x} $$

Part (b)

Differentiate the equation from part (a) with respect to r: $$ a \frac{\partial}{\partial r}\left(\frac{\partial^{2}}{\partial^{2} x}\left(e^{r x}\right)\right)+b \frac{\partial}{\partial r}\left(\frac{\partial}{\partial x}\left(e^{r x}\right)\right)+c\left(x e^{r x}\right)=\left[2+\left(r-r_{1}\right) x\right] a\left(r-r_{1}\right) e^{r x} $$ Compute the partial derivatives \(\frac{\partial}{\partial r}\left(\frac{\partial^{2}}{\partial^{2} x}\left(e^{r x}\right)\right)\) and \(\frac{\partial}{\partial r}\left(\frac{\partial}{\partial x}\left(e^{r x}\right)\right)\): $$ \frac{\partial}{\partial r}\left(\frac{\partial^{2}}{\partial^{2} x}\left(e^{r x}\right)\right) = 2rxe^{r x} $$ $$ \frac{\partial}{\partial r}\left(\frac{\partial}{\partial x}\left(e^{r x}\right)\right) = xe^{r x} $$ Substitute these partial derivatives into the equation: $$ a(2rxe^{r x})+b(xe^{r x})+c\left(x e^{r x}\right)=\left[2+\left(r-r_{1}\right) x\right] a\left(r-r_{1}\right) e^{r x} $$

Part (c)

Reverse the order of the partial differentiations in the first two terms on the left side of the equation above: $$ a \frac{\partial^{2}}{\partial x^{2}}\left(x e^{r x}\right)+b\frac{\partial}{\partial x}\left(x e^{r x}\right)+c\left(x e^{r x}\right)=\left[2+\left(r-r_{1}\right) x\right] a\left(r-r_{1}\right) e^{r x} $$

Part (d)

Set \(r=r_{1}\) in the equations from part (b) and part (c) to find the solutions y1 and y2: For part (b) equation: $$ a(2r_{1}xe^{r_{1} x})+b(xe^{r_{1} x})+c\left(x e^{r_{1} x}\right)=2 a e^{r_{1} x} $$ Divide both sides by \(e^{r_{1} x}\) and observe that the left side becomes the original equation. Therefore, \(y_{1}=e^{r_{1} x}\) is a solution. For part (c) equation: $$ a \frac{\partial^{2}}{\partial x^{2}}\left(x e^{r_{1} x}\right)+b\frac{\partial}{\partial x}\left(x e^{r_{1} x}\right)+c\left(x e^{r_{1} x}\right)=2 a e^{r_{1} x} $$ Divide both sides by \(x e^{r_{1} x}\). Then, the left side becomes the original equation. Therefore, \(y_{2}=x e^{r_{1} x}\) is a solution.

Key concepts

Partial Differentiation

Understanding partial differentiation is crucial for solving differential equations, especially when dealing with multiple variables. Partial differentiation involves taking the derivative of a function with respect to one variable while keeping other variables constant.

In the context of solving a homogeneous second order linear differential equation, partial differentiation allows us to explore the changes in the function as we slightly vary one of the variables. For instance, the function given in the exercise, \(e^{rx}\), is treated as a function of both \(x\) and \(r\). When we compute \(\frac{\text{partial}}{\text{partial} x}\left(e^{rx}\right)\), we differentiate with respect to \(x\) while treating \(r\) as a constant. This concept is pivotal for finding solutions to our differential equation, as it shows the relation between the function and its growth rate with respect to one variable at a time.

Second Order Linear Differential Equation

A second order linear differential equation is characterized by derivatives up to the second order, and it is crucial in modelling physical systems like oscillations and wave propagation.

The general form of such an equation is \(a y'' + b y' + c y = 0\), where \(a, b,\) and \(c\) are constants, and the primes denote derivatives with respect to \(x\). The characteristic equation is used to solve these equations, which in our exercise takes the form of a quadratic equation whose roots determine the behavior of the solutions. It is essential for students to grasp how the characteristic equation relates to the solutions of differential equations and how the nature of the roots affects the form of the general solution.

Repeated Real Roots

The presence of repeated real roots in the characteristic equation of a differential equation significantly affects the form of the general solution.

In the given exercise, we have a repeated root \(r_1\), which implies that the characteristic equation has the same root for both solutions of our second order differential equation. Ordinarily, each distinct root would generate an independent solution to the equation, but repeated roots introduce a slightly more complex scenario. In such cases, the general solution includes terms like \(y_1 = e^{r_1x}\) and \(y_2 = xe^{r_1x}\), with the second solution having an additional multiplicative factor of \(x\). This ensures that both solutions are linearly independent, fulfilling the requirement for a complete solution set to our homogeneous differential equation.

Homogeneous Differential Equations

Homogeneous differential equations are a class of differential equations where the right-hand side is zero, formally written as \(a y'' + b y' + c y = 0\) where \(y\) is the dependent variable and \(x\) is the independent variable.

In our exercise, the differential equation is homogeneous. Such equations are significant because their solutions can be linearly combined to produce more solutions—a principle of superposition. Homogeneous equations are often easier to solve compared to their non-homogeneous counterparts. The solutions, as in our exercise, usually rely on exploiting the characteristics of exponential functions and their derivatives, which can be done through methods like solving the characteristic equation or using the method of undetermined coefficients for more complex cases.