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− | Let's do an example to make it clear. Consider the second-order ODE: <math>\frac{d^2y}{dt^2}+(cos^2t)\frac{dy}{dt}+y= | + | Let's do an example to make it clear. Consider the second-order ODE: <math>\frac{d^2y}{dt^2}+(cos^2t)\frac{dy}{dt}+y=0</math>. We substitute the first-order differential term <math>\frac{dy}{dt}</math> by a new variable <math>x</math>. In order to make the final expression look more like a system, we rename <math>y=x_1</math>, <math>\frac{dy}{dt}=x=x_2</math>. |

Here comes a "magic" derivative conversion, <math>\frac{dx_1}{dt}=\frac{dy}{dt}=x_2</math>, | Here comes a "magic" derivative conversion, <math>\frac{dx_1}{dt}=\frac{dy}{dt}=x_2</math>, | ||

− | <math>\frac{dx_2}{dt}=\frac{d\frac{dy}{dt}}{dt}=\frac{d^2y}{dt^2}</math>. | + | <math>\frac{dx_2}{dt}=\frac{d\frac{dy}{dt}}{dt}=\frac{d^2y}{dt^2}=-cos^2t \frac{dy}{dt}-y=(-cos^2t)x_2-x_1</math> by rearranging the initial ODE. |

− | + | Therefore, the converted system of ODEs is | |

+ | <math>\frac{dx_1}{dt}=x_2</math>, | ||

− | ''' <big><big><big> 5. | + | <math>\frac{dx_2}{dt}=-x_1-(cos^2t)x_2</math>. |

+ | |||

+ | We write it into matrix form to make it more clear: | ||

+ | |||

+ | <math>\begin{bmatrix} | ||

+ | \frac{dx_1}{dt}\\ | ||

+ | \frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix} | ||

+ | 0 & 1 \\ | ||

+ | -1 & cos^2t \end{bmatrix} \begin{bmatrix} | ||

+ | x_1 \\ | ||

+ | x_2 \end{bmatrix} </math>. | ||

+ | |||

+ | |||

+ | In this example, the term without a differential term <math>\frac{dy}{dt}</math> is <math>0</math>. If we consider the differential term as an algebraic variable, then the second-order polynomial- the parabola will pass the origin. In algebraic equations, this guarantees the ability of the polynomial of factorisation. Similarly, in ODEs, no zero-powered differential term guarantees the ability of the ODE of matrix conversion. | ||

+ | |||

+ | Imagine an algebraic equation in higher orders with a non-zero constant (e.g. <math>ax^2+bx+c=0</math>, <math>c≠0</math>), which means the same with here, the term without the differential term <math>\frac{dy}{dt}</math> is non-zero. If we change the right-hand side of the above example to any non-zero term without differential term or the variable itself, the converted system of first-order ODEs will be unavailable to be written in a matrix form, but it is still good enough for further operations. | ||

+ | |||

+ | Consider <math>\frac{d^2y}{dt^2}+(cos^2t)\frac{dy}{dt}+y=3t</math>. Follow the above operations to convert it into a ODE system, we will have | ||

+ | |||

+ | <math>\frac{dx_1}{dt}=x_2</math>, | ||

+ | |||

+ | <math>\frac{dx_2}{dt}=-x_1-(cos^2t)x_2+3t</math>. | ||

+ | |||

+ | Done. We can start solving the system by various methods now. </font> | ||

+ | |||

+ | |||

+ | ''' <big><big><big> 5.2 References </big></big></big> ''' | ||

<font size="3px"> Institute of Natural and Mathematical Science, Massey University. (2017). ''160.204 Differential Equations I: Course materials.'' Auckland, New Zealand. | <font size="3px"> Institute of Natural and Mathematical Science, Massey University. (2017). ''160.204 Differential Equations I: Course materials.'' Auckland, New Zealand. |

## Latest revision as of 22:15, 21 November 2017

## Introduction to ODEs in Higher Orders

### 5.0 Abstract

In last tutorial we looked at three basic methods to solve differential equations in the first order. In a linear equation, we can switch the variable $ x $ to a higher order, like $ x^2 $, $ x^3 $, ..., $ x^n $ to obtain higher-ordered equations. Similarly, the differential term $ \frac{dy}{dx} $ can also be switched as $ \frac{d^2y}{dx^2} $, $ \frac{d^3y}{dx^3} $, ..., $ \frac{d^ny}{dx^n} $. This gives us the basic idea of differential equations in higher orders, the most general form for which is like $ f_n(t)\frac{d^ny}{dt^n}+f_{n-1}\frac{d^{n-1}y}{dt^{n-1}}+...+f_1(t)\frac{dy}{dt}+f_0(t)y=g(t) $, where $ n $ is the order.

A direct idea to deal with ODEs in higher orders is to convert them into a linear system of ODEs, which is what we are focusing at in this short tutorial. Other solutions like Laplace transforms, variation of constants and Cauchy-Euler equations will come up later.

** 5.1 Converting to Linear Systems **

An important idea is substitution, similar as the substitution method for integration. We substitute a first-order differential term by a new variable, to reduce the order and introduce the new variable to form a system of first-order ODEs.

Let's do an example to make it clear. Consider the second-order ODE: $ \frac{d^2y}{dt^2}+(cos^2t)\frac{dy}{dt}+y=0 $. We substitute the first-order differential term $ \frac{dy}{dt} $ by a new variable $ x $. In order to make the final expression look more like a system, we rename $ y=x_1 $, $ \frac{dy}{dt}=x=x_2 $.

Here comes a "magic" derivative conversion, $ \frac{dx_1}{dt}=\frac{dy}{dt}=x_2 $,

$ \frac{dx_2}{dt}=\frac{d\frac{dy}{dt}}{dt}=\frac{d^2y}{dt^2}=-cos^2t \frac{dy}{dt}-y=(-cos^2t)x_2-x_1 $ by rearranging the initial ODE.

Therefore, the converted system of ODEs is

$ \frac{dx_1}{dt}=x_2 $,

$ \frac{dx_2}{dt}=-x_1-(cos^2t)x_2 $.

We write it into matrix form to make it more clear:

$ \begin{bmatrix} \frac{dx_1}{dt}\\ \frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & cos^2t \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $.

In this example, the term without a differential term $ \frac{dy}{dt} $ is $ 0 $. If we consider the differential term as an algebraic variable, then the second-order polynomial- the parabola will pass the origin. In algebraic equations, this guarantees the ability of the polynomial of factorisation. Similarly, in ODEs, no zero-powered differential term guarantees the ability of the ODE of matrix conversion.

Imagine an algebraic equation in higher orders with a non-zero constant (e.g. $ ax^2+bx+c=0 $, $ c≠0 $), which means the same with here, the term without the differential term $ \frac{dy}{dt} $ is non-zero. If we change the right-hand side of the above example to any non-zero term without differential term or the variable itself, the converted system of first-order ODEs will be unavailable to be written in a matrix form, but it is still good enough for further operations.

Consider $ \frac{d^2y}{dt^2}+(cos^2t)\frac{dy}{dt}+y=3t $. Follow the above operations to convert it into a ODE system, we will have

$ \frac{dx_1}{dt}=x_2 $,

$ \frac{dx_2}{dt}=-x_1-(cos^2t)x_2+3t $.

Done. We can start solving the system by various methods now.

** 5.2 References **

Institute of Natural and Mathematical Science, Massey University. (2017). *160.204 Differential Equations I: Course materials.* Auckland, New Zealand.

Robinson, J. C. (2003). *An introduction to ordinary differential equations.* New York, NY., USA: Cambridge University Press.

Schaft, A. J. (1986). On Realisation of Nonlinear Systems Described by Higher-Order Differential Equations. *Mathematical Systems Theory, 19* (1), p.239-275. DOI: https://doi.org/10.1007/BF01704916.