Linear ODEs

Differential equations: Introduction to Differential equations

Linear ODEs

We will now pay attention the following type of differential equation.

Let $\varphi(t,y,y',y'',\ldots)=0$ be an ordinary differential equation in the unknown function $y$ of the single variable $t$ .

The differential equation is called linear if the function $\varphi$ can be written as a sum of terms that are either a function of $t$ or a product of a function of $t$ with $y$ or a (higher) derivative of $y$ . Otherwise, the ODE is called nonlinear.
A linear differential equation is called homogeneous if each term in the equation is a multiple of $y$ or of a derivative of $y$ . Otherwise the linear ODE is called non-homogeneous or inhomogeneous.

The first-order ODE $\frac{\dd y}{\dd t}=t\cdot y$ can be written as $\varphi(t,y,y')=0$ with $\varphi(u,v,w)=u\cdot v-w$ . It is a homogeneous linear differential equation of order 1 and degree 1.
The ODEs $y'-y^2=0\quad\text{and}\quad y'-y-1=0$ are examples of first-order differential equations of first degree.
The ODE $(y'')^3-y^2=0$ is an example of a second-order differential equation of degree 3.
The ODE $(y'')^3-y\cdot y'=0$ is an example of a non-linear second-order differential equation of degree 3.

A linear differential equation of order $3$ is of the form

$a_3(t)\cdot y'''+a_2(t)\cdot y''+a_1(t)\cdot y'+a_0(t)\cdot y +b(t)=0$

where $a_3$ , $a_2$ , $a_1$ , $a_0$ , and $b$ are functions. The ODE is homogeneous if and only if $b=0$ .

The following statement explains the term "linear" for an ODE and helps solving them.

Linear structure of linear ODEs

Let $a_0(t),a_1(t),\ldots,a_n(t)$ and $b(t)$ be continuous functions with $a_n(t)\ne0$ such that

$a_n(t)\cdot y^{(n)}+\cdots+a_2(t)\cdot y''+a_1(t)\cdot y'+a_0(t)\cdot y +b(t)=0$

is a linear differential equation of order $n$ . This equation is homogeneous if and only if $b(t)=0$ .

Suppose that $y_{\text{part}}$ is a solution of this equation. Then every other solution has the form

$u+y_{\text{part}}$ where $u$ is a solution of the corresponding homogeneous ODE (that is, the equation with $b(t)$ replaced by $0$ ).

We call the solution $y_{\text{part}}$ a particular solution of the ODE.

If $b(t)=0$ , then the set of solutions is linear in the following sense:

The constant function $u=0$ is a solution.
If $\alpha$ and $\beta$ real numbers and $u$ and $v$ are solutions of the homogeneous ODE, then so is $\alpha\cdot u+\beta\cdot v$ .

For those who are familiar with linear algebra: linearity of the homogeneous equation

$a_n(t)\cdot y^{(n)}+\cdots+a_2(t)\cdot y''+a_1(t)\cdot y'+a_0(t)\cdot y =0$

means that the set of solutions is a linear subspace of the vector space of all differentiable functions. In general, this vector space is $n$ -dimensional, so each solution is a unique linear combination of $n$ different solutions.

The solutions of the original (possibly inhomogeneous) equation form an affine subspace with position vector the private solution $y_{\text{part}}$ .

If the initial ODE is homogeneous, then we take $y_{\text{part}}=0$ to be the particular solution.

We give the derivation of the key steps in case $n=2$ . The general case is not really more difficult, but leads to more writing. The linear properties are a consequence of the following calculation for the left-hand side of the equation for $y=\alpha \cdot u+\beta\cdot v$ for two real numbers $\alpha$ , $\beta$ and two functions $u$ , $v$ :

$\begin{array}{rcl} {\small \phantom{xx}}&&a_2(t)\cdot y''+a_1(t)\cdot y'+a_0(t)\cdot y \\ &&= a_2(t)\cdot \left(\alpha \cdot u''+\beta\cdot v''\right)\\ &&\phantom{xx}+a_1(t)\cdot\left(\alpha \cdot u'+\beta\cdot v'\right)+a_0(t)\cdot \left(\alpha \cdot u+\beta\cdot v\right) +b(t)\\ &&= a_2(t)\cdot \alpha \cdot u''+a_2(t)\cdot \beta\cdot v''+a_1(t)\cdot\alpha \cdot u'+a_1(t)\cdot\beta\cdot v' \\ &&\phantom{xx}+a_0(t)\cdot \alpha \cdot u+a_0(t)\cdot \beta\cdot v +b(t)\\ &&= a_2(t)\cdot \alpha \cdot u''+a_1(t)\cdot\alpha \cdot u'+a_0(t)\cdot \alpha \cdot u\\ &&\phantom{xx}+a_2(t)\cdot \beta\cdot v''+a_1(t)\cdot\beta\cdot v'+a_0(t)\cdot \beta\cdot v +b(t)\\ &&= \alpha\cdot\left(a_2(t) \cdot u''+a_1(t)\cdot u'+a_0(t) \cdot u\right)\\ &&\phantom{xx}+\beta\cdot\left(a_2(t)\cdot v''+a_1(t)\cdot v'+a_0(t)\cdot v \right)+b(t) \end{array}$

If $\alpha=\beta=1$ and $v=y_{\text{part}}$ , the bottom line is equal to $0$ . This shows that $y= u+y_{\text{part}}$ is a solution of the ODE if and only if $u$ is a solution of the homogeneous equation.

If $b(t)=0$ and $u$ and $v$ are both solutions of the homogeneous equation, then the two bottom lines are equal to $0$ so $y=\alpha \cdot u+\beta\cdot v$ is a solution of the homogeneous equation.

In order to find a particular solution $y_{\text{part}}$ of a linear ODE, we can make additional assumptions that simplify the search. We must then assess the compliance of the solutions to those assumptions. Below are some examples. For the solution of the corresponding homogeneous equation, at least in the case of order $1$ and in special cases for order $2$ , we will present a general method.

Below we present the differential equation of Van der Pol:
${{\dd ^2}\over{\dd x^2}}\cdot y-\mu\cdot \left(1-y^2\right)\cdot \left({{\dd }\over{\dd x}}\cdot y\right)+y=0$
It is a second-order equation of degree $1$ . Is this ODE linear?

no

We replace $y'$ by $u$ and $y''$ by $v$ in the left side of the equation
$-\mu\cdot u\cdot \left(1-y^2\right)+y+v$
This is not a linear expression in the variable $y$ . Therefore, the ODE is not linear.

New example