Difference between revisions of "Introduction to ODEs in Higher Orders" - Rhea

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<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} $.

In this example, the term without a differential term $ \frac{dy}{dt} $ is $ 0 $. If we consider the differential term as a normal variable, then the second-order polynomial- the parabola will pass the origin. In normal 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 a normal 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 $,