Higher Order Linear

Introduction

Linear Differential Equaiton

We say that a differential equation is linear if it is a linear polynomial in the function and its derivatives. $$a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\dots+a_1(x)y'+a_0(x)y=g(x)$$

Homogeneous Case

We can call a differential equation homogeoneous if g(x)=0

Superposition Principle

If \(y_{1}(x)\) and \(y_{2}(x)\) are two solutions of a homogeneous linear differential equation, then any linear combination \(c_1y_1(x)+c_2y_2(x)\) is also a solution.

Characteristic Equation

Let us consider a second order homogeneous linear differential equation $$ay''+by'+cy=0$$ Since the function and its derivatives must cancel each other out, we might guess that the answer is an exponential. $$y=e^{mx}$$ We have the following derivatives for y: $$y'=me^{mx}$$ $$y''=m^2e^{mx}$$ Plugging in to our differential equation: $$am^2e^{mx}+bme^{mx}+ce^{mx}=0$$ $$(am^2+bm+c)e^{mx}=0$$ If the product of two factors is 0, then at least one of the factors must be 0. Because \(e^{mx}\) will never be 0, we need not consider it for finding roots. We can then focus on the equation $$am^2+bm+c=0$$ Which is a simple polynomial of m. We can then find the values of m for which the equation is 0, by factoring, quadratic equation, or completing the square.

Let \(m_1\) and \(m_2\) be the roots of the polynomial.
If \(m_1\neq m_2\), we have the following solution $$y=c_1e^{m_1x}+c_2e^{m_2x}$$ If \(m_1=m_2\), or we have repeated roots in the higher order case, the repeats will have an additional factor of x for the number repeat it is. $$y=c_1e^{m_1x}+c_2xe^{m_1x}$$

Complex Roots

But what if we have complex roots?
\(e^{mx}\) is still valid for complex m. Euler's formula gives the relationship for a complex exponent. $$e^{ix}=\cos x+i \sin(x)$$ Another relevant formula is that of de Moivre's which states $$(\cos x+i\sin x)^n=\cos nx+i\sin nx$$ The real part of an exponent can be factored out, letting us get $$e^{(a+bi)x}=e^{ax}\left[cos(bx)+i\sin(bx)\right]$$ Since actually do not care whether the constants in front of solutions are real or complex, we get the solutions $$y=e^{ax}\left[c_1\cos(bx)+c_2\sin(bx)\right]$$ as the solution if the root is m=a+bi

Python Code

Cauchy-Euler Equations

Cauchy-Euler equations are a class of linear differential equations that have variable coefficients. The general form of a Cauchy-Euler equation of order n is given by: $$x^ny^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\dots+a_1xy'+a_0y=0$$

Characteristic Equation

While we previously assumed an exponential solution for regular homogeneous equations, for Cauchy-Euler equations we assume a solution of the form $$y=x^m$$ Plugging this in gives us the characteristic equation $$m(m-1)(m-2)\dots(m-(n-1))+a_{n-1}m(m-1)\dots(m-2)+\dots+a_0=0$$ We solve the polynomial for m similar to before.

Solutions

For distinct real roots, the general solution is $$y(x)=c_1x^{m_1}+\dots+c_nx^{m_n}$$ If a root m is repeated k times, the corresponding terms gain powers of \(\ln(x)\) $$x^m,x^m\ln(x),x^m(\ln(x))^2\dots,x^m(\ln(x))^{k-1}$$ For complex roots \(m=\alpha\pm i\beta\), the corresponding terms in the general solution are $$x^\alpha\cos(\beta\ln(x))\text{ and }x^\alpha\sin(\beta\ln(x))$$

Nonhomogeneous

But what if \(g(x)\neq 0\)?

Python Code There are multiple ways to solve these sorts of equations.

Method of Undetermined Coefficients

Cases

This can be used when g is a polynomial, exponential, sine, and/or cosine function.

Steps

1. Guess the form
2. Differentiate the guess
3. Substitute into the differential equation
4. Equate coefficients

Overlaps

If the guessed form is already a solution to the homogeneous version of the differential equation, multiply the guess by a factor of t.

Variation of Parameters

Variation of parameters is a more widely applicable method compared to Method of Undetermined Coefficients.

Steps

1. Determine homogeneous solutions
2. Construct Wronskian matrix
3. Get coefficients for homogeneous solution

Determine homogeneous solution

Use previous methods to solve the case where g(x)=0

Construct Wronskian

For an equation with two homogeneous solutions, the Wronskian is given by: $$W=det\begin{bmatrix} y_1 & y_2\\ y_1' &y_2' \end{bmatrix}$$ This simplifies to \(y_1y_2'-y_2y_1'\)

Get Coefficients

The derivatives of the coefficients are given by $$u_1'=\frac{-y_2g(t)}{W}$$ $$u_2'=\frac{y_1g(t)}{W}$$ Upon integrating, this gives the solution $$y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$$

Applications

Spring

Exercises

1. Solve the Cauchy-Euler equation:
   1. \(x^2y''+xy'-y=0\)