Question

(a) Consider the equation \(y^{\prime \prime}+2 a y^{\prime}+a^{2} y=0 .\) Show that the roots of the characteristic equation are \(r_{1}=r_{2}=-a,\) so that one solution of the equation is \(e^{-a t}\). (b) Use Abel's formula [Eq. (8) of Section \(3.3]\) to show that the Wronskian of any two solutions of the given equation is $$ W(t)=y_{1}(t) y_{2}^{\prime}(t)-y_{1}^{\prime}(t) y_{2}(t)=c_{1} e^{-2 a t} $$ where \(c_{1}\) is a constant. (c) Let \(y_{1}(t)=e^{-a t}\) and use the result of part (b) to show that a second solution is \(y_{2}(t)=t e^{-a t} .\)

Short answer

Question: Show that the roots of the given homogeneous linear differential equation's characteristic equation are \(r_1 = r_2 = -a\), and find the Wronskian and a second linearly independent solution for the given differential equation: $$y'' + 2ay' + a^2y=0$$ Short Answer: The roots of the characteristic equation are \(r_1 = r_2 = -a\). The Wronskian is given by \(W(t) = c_1 e^{-2at}\), where \(c_1\) is a constant. A second linearly independent solution is \(y_2(t)=te^{-at}\).

Step-by-step solution

Part (a): Finding the roots of the characteristic equation

To solve the given homogeneous linear differential equation, we first have to find its characteristic equation. That is given by: $$r^2 + 2ar + a^2 = 0$$ Now, we have to find the roots, \(r_1\) and \(r_2\), of the characteristic equation using the quadratic formula: $$r_{1,2}=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$ For our equation, we have $$A = 1, B = 2a, \mathrm{and} C = a^2$$. Plugging in the values: $$r_{1,2}=\frac{-(2a)\pm\sqrt{(2a)^2-4(1)(a^2)}}{2(1)}$$ Upon simplification, we find that the discriminant inside the square root is zero. $$\Rightarrow r_{1,2}=-a$$ Since \(r_1 = r_2 = -a\), one solution of the given differential equation is \(y_1(t) = e^{-at}\).

Part (b): Using Abel's Formula

Recall Abel's formula for a homogeneous linear second-order differential equation of the form: $$y^{\prime\prime} + p(t)y^{\prime} + q(t)y = 0$$ The Wronskian of any two solutions is given by: $$W(t) = \frac{c}{e^{\int p(t)\, dt}}$$ In our given equation, the function \(p(t)=2a\) and \(q(t) = a^2\). Now, we will find the Wronskian: $$W(t) = \frac{c}{e^{\int 2a\, dt}}$$ $$W(t) = \frac{c}{e^{2at}} = c_1 e^{-2at}$$ where \(c_1\) is a constant.

Part (c): Finding the second linearly independent solution

Given \(y_1(t)=e^{-at}\), we can use the result from part (b) to find a second linearly independent solution. We know that: $$W(t) = y_1(t) y_2^{\prime}(t)-y_1^{\prime}(t) y_2(t) = c_1 e^{-2 a t}$$ Differentiating \(y_1(t)\), we get $$y_1^{\prime}(t)=-ae^{-at}$$ Now let's plug in \(y_1(t)\) and \(y_1^{\prime}(t)\) into Wronskian formula. $$c_1 e^{-2at} = \left(e^{-at}\right)\left(y_2^{\prime}(t)\right) - \left(-ae^{-at}\right)\left(y_2(t)\right)$$ Dividing both sides by \(c_1 e^{-at}\), we get $$y_2^{\prime}(t) - a y_2(t) = -a e^{-at}$$ Now looking at the right side of the equation, we observe that multiplying \(e^{-at}\) by \(t\) gives us a second solution, \(y_2(t) = te^{-at}\). Thus, the second linearly independent solution is given by: $$y_2(t)=te^{-at}$$

Key concepts

Characteristic Equation

In the realm of differential equations, the characteristic equation is a pivotal tool for solving homogeneous linear differential equations. It is a concise representation that unlocks the structure of the equation's solutions. For example, when examining the equation \( y'' + 2ay' + a^2y = 0 \), we must first identify its characteristic equation: \( r^2 + 2ar + a^2 = 0 \).

To uncover the roots, which are the values of \( r \) that satisfy the equation, we often employ the quadratic formula:\[ r_{1,2} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]By substituting the coefficients into the formula, we deduce that both roots are equal \( r_1 = r_2 = -a \), indicative of a repeated root scenario. This immediately gives us a solution to the original differential equation in the form \( y_1(t) = e^{-at} \), an exponential function that captures the equation's deterioration over time.

Wronskian

The Wronskian is an expression of great significance in the study of differential equations, serving as a detective of sorts by probing into the behavior of solutions. It is defined as the determinant of a matrix that includes the functions and their derivatives. Specifically, for two functions \( y_1(t) \) and \( y_2(t) \), the Wronskian is \[ W(t) = y_1(t)y_2'(t) - y_1'(t)y_2(t) \].

By invoking Abel's formula, we can appreciate its underlying pattern. For the differential equation \( y'' + 2ay' + a^2y = 0 \), we can utilize Abel's formula to narrate that the Wronskian takes the form \( W(t) = c_1e^{-2at} \), where \( c_1 \) is a constant. This reveals how the Wronskian, much like a rhythm, pulsates at a rate governed by the coefficient \( 2a \) from the differential equation.

Abel's Formula

Delving into the heart of homogeneous linear second-order differential equations, we encounter Abel's formula, a bridge connecting the coefficients of the equation to the Wronskian of its solutions. The formula has a deceptively simple beauty:\[ W(t) = \frac{c}{e^{\int p(t) dt}} \]Here, \( p(t) \) is a function drawn from the standard form of the equation \( y'' + p(t)y' + q(t)y = 0 \).

When applied to our example, with \( p(t) = 2a \), Abel's formula tells a tale of decay, mirroring the exponential decrease of the Wronskian, \( W(t) = c_1e^{-2at} \). The beauty rests in its ability to distill the essence of the equation's behavior into an elegant statement about the interplay between the solutions.

Linearly Independent Solutions

Navigating through the terrain of differential equations, the notion of linearly independent solutions becomes as essential as a compass for an explorer. This concept asserts that each solution in a set is novel, contributing a distinct direction to the space of all solutions. For our equation \( y'' + 2ay' + a^2y = 0 \), the first solution is given by \( y_1(t) = e^{-at} \), but to sketch the full landscape, another unique pathway must be found.

By wielding the Wronskian and its pivotal role in the narrative, we confirm that a second, distinct solution exists in the form \( y_2(t) = te^{-at} \). This function, elegantly bisected by \( t \), ensures that our two solutions provide separate trails across the solution space, much like orthogonal vectors define dimensions in space. Together, these linearly-independent solutions form the complete solution to our differential equation, each playing its part in the ensemble that describes the system's evolution.