ch-1-introduction

What Is a Differential Equation?

A differential equation, as the words suggest, is an equation that involves derivatives or differentials of an unknown function.

Therefore,

\[\frac{d^2 y}{dx^2}+5 y=0\]\[\left(x \frac{dy}{dx}-y\right)^2=(x+y)^2\]\[\left\{\begin{aligned} &\frac{dy}{dt}-3\frac{dx}{dt}=0\ &\frac{dy}{dt}+5\frac{dx}{dt}=e^{-t} \end{aligned}\right.\]\[\frac{\frac{dy}{dx}}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}=1\]\[x^2y\, dx+x^3y^2\, dy=0\]\[x\frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=z\]

are examples of differential equations.

Equations that involve a single independent variable (and therefore only ordinary derivatives occur) are known as ordinary differential equations (ODEs).

When there are more than one independent variable, partial derivatives are involved and the equation is called a partial differential equation (PDE).

Notation: The expressions $y', y'', y''', y^{(4)}, \cdots, y^{(n)}$ represent the first, second, third, fourth, $\cdots$, $n$th derivatives of $y$ with respect to the independent variable under consideration. For example, $y'''$ means $\dfrac{d^3 y}{dx^3}$ if the independent variable is $x$. In physics, when the independent variable is time $t$, dots are sometimes used instead of primes. Therefore, $\dot{y}=\dfrac{dy}{dt}$ and $\ddot{y}=\dfrac{d^2y}{dt^2}$.

The equation $\displaystyle{\frac{dy}{dx}=5x+4y}$ or $y'=5x+4y$ in which $y$ is regarded as the unknown function of $x$ is an ODE.
The equation $\displaystyle{y+\left(\frac{d^3 y}{dx^3}\right)^2=\sin x}$ which may also be written as $y+(y''')^2=\sin x$ is an ODE.
The equation $\displaystyle{\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}}=f(x,y)$ in which $V$ is an unknown function of $x$ and $y$ and $f$ is a given function is a PDE.
The equation $\displaystyle{\frac{\partial u}{\partial t}=\frac{\partial ^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}}$ in which $u$ is the unknown function of $x$, $y$, and $t$ is a PDE.
The equation $\dfrac{\partial u}{\partial t}+u\dfrac{\partial u}{\partial x}=\nu \dfrac{\partial^2 u}{\partial x^2}$ in which $\nu$ is a constant and $u$ is the unknown function of $x$ and $t$ is a PDE. This equation is called Burger's equation.

The Order of a Differential Equation

The order of a differential equation is the order of the highest derivative (or partial derivative) occurring in the equation.

For example,

\[5\frac{d^4y}{dx^4}+4y=0\]

is an ordinary differential equation of order 4, and

\[\frac{\partial u}{\partial t}=\rho \frac{\partial^2 u}{\partial x^2}\]

is a partial differential equation of order 2.

An $n$th order ODE can be expressed as

\[F\left(x,y,y',y'',\cdots,y^{(n)}\right)=0 \tag{i}\]

where $F$ is some function of $n+2$ variables: $x$, $y$, $y'$, $y''$, $\dots$, $y^{(n)}$. If we can solve this relation for the highest derivative, we can rewrite it in the form

\[y^{(n)}=G\left(x,y,y',\cdots,y^{(n-1)}\right),\tag{ii}\]

where $G$ is another function.

A second-order PDE for $u(x,t)$ can be expressed as

\[H\left(x,t,u,\frac{\partial u}{\partial x},\frac{\partial u}{\partial t},\frac{\partial^2 u}{\partial x^2},\frac{\partial^2 u}{\partial x \partial t},\frac{\partial^2 u}{\partial t^2}\right)=0\]

Degree of a Differential Equation

The degree of a differential equation is the exponent of its highest-ordered derivative. To find the degree, we must rationalize the equation and clear fractions.

Example: In the differential equation $ (y''')^2+y' y^2+(y'')^4=x^5 $ the degree is 2 because the exponent of the highest-ordered derivative, $y'''$, is 2.

Linear vs. nonlinear, homogeneous vs. non-homogeneous

The ordinary differential equation $F(x, y, y', \dots, y^{(n)})=0$ is said to be linear if the dependent variable and its derivatives appear only to the first degree. In other words, a linear ordinary differential equation of order $n$ is an equation of the form \[a_n(x)\frac{d^n y}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y(x)=b(x) \tag{iii} \] where the coefficients $a_n(x), \dots, a_0(x)$, and $b(x)$ are given functions of the independent variable $x$.

The distinguishing characteristic of a linear ordinary differential equation (linear ODE) is:

$y, \dfrac{dy}{dx}, \cdots, \dfrac{d^n y}{dx}$ all have exponent one.
There is no product or nonlinear function of $y$ or its derivatives.

Also notice that:

If $F(x, y, y', \dots, y^{(n)})=0$ is a linear differential equation, $F$ is a linear function of the dependent variable and its derivatives ($y$, $y'$, $\dots$, and $y^{(n)}$). However, $F$ does not have to be a linear function of $x$ for the differential equation to be linear.

Some examples on linear and nonlinear differential equations

The equation
\[\frac{d^2y}{dt^2}-\mu(1-y^2)\frac{dy}{dt}+\nu\, y=0\]
where $\mu$ and $\nu$ are two constants is a nonlinear differential equation because this equation involves a product of $y^2$ (a nonlinear term) and $\frac{dy}{dt}$.
The differential equation
\[\frac{d^2y}{dt^2}+\frac{g}{L} \sin(y)=0\]
where $g$ and $L$ are two constants is nonlinear because of the presence of $\sin(y)$.
The equation
\[\frac{d^2y}{dt^2}+\sin t \frac{dy}{dt}=\cos t\]
is a linear equation as it is in the form of Equation (iii).

If $b(x)\equiv 0$, we say that Equation (iii) is homogeneous; otherwise it is non-homogeneous.

A partial differential equation (PDE) is said to be linear if the equation if no power or product of the unknown function and its partial derivatives are present after rationalizing the equation and clearing fractions. For examples, a linear second-order partial differential equation for $u(x,t)$ has the form

\[ a_1(x,t) \frac{\partial^2 u}{\partial x^2}+a_2(x,t) \frac{\partial^2 u}{\partial x \partial t}+a_3(x,t)\frac{\partial^2 u}{\partial t^2}+a_4(x,t)\frac{\partial u}{\partial x}+a_5(x,t) \frac{\partial u}{\partial t}+a_6(x,t) u=b(x,t) \]

where $a_1(x,t),\dots , a_6(x,t)$, and $b(x,t)$ are some given functions.

In general, a linear partial differential equation for $u(x,t)$ has the form

\[\sum_{n=0}^N\ \sum_{m=0}^M a_{mn}(x,t) \frac{\partial^{m+n} u}{\partial x^n \partial t^m} =b(x,t)\tag{iv} \]

where $a_{mn}(x,t)$ and $b(x,t)$ are given functions of $x$ and $t$, and $M$ and $N$ are fixed positive integers. We define $\dfrac{\partial^0 u}{\partial x^0\partial t^0}= u$.

Equation (iv) is homogeneous if $b(x,t)\equiv0$.

Explicit and Implicit Solutions of a Differential Equation

Consider a general differential equation of order $n$ that is described by \[F(x, y, y', \dots, y^{(n)})=0.\tag{v}\] We say a function $\phi(x)$ is a solution of the above differential equation in a given interval $I$ if when we substitute $\phi(x)$ into the equation (i.e. replacing $y$ by $\phi(x)$, $y'$ by $\phi'(x)$, $\dots$, $y^{(n)}$ by $\phi^{(n)}(x)$), the equation reduces to an identity at every point of $I$: \[F[x,\phi(x), \phi'(x),\dots, \phi^{(n)}(x)]=0.\] Such a function $\phi(x)$ is said to satisfy the equation in $I$ and is called a solution of the above equation. The graph of $\phi$ is called a solution curve.

For example, $y=\sin x$ is a solution of

\[y^{\prime\prime}+y=0\tag{vi}\]

because

$ (\sin x)^{\prime\prime}+\sin x=0 \quad (-\infty < x < \infty) $

Similarly, we can show $y=\cos x$ is also a solution of this differential equation. In fact, any function of the form

\[y=c_1 \sin x+c_2 \cos x,\tag{vii}\]

where $c_1$ and $c_2$ are two arbitrary constants is also a solution. This means that the equation $y^{\prime\prime}+y=0$ has infinitely many solutions. However, not all differential equations have infinitely many solutions. A given differential equation may have a unique solution, multiple solutions, or no solution at all.

We call the linear combination (vii) the general solution of the equation (vi) as every single solution to this equation is of this form.

We have already shown that the equation $y''+y=0$ has infinitely many solutions because every value assigned to $c_1$ and $c_2$ in equation (vii) generates a different solution.
The differential equation $(y'')^2+y^2=0$ has only one solution, which is $y\equiv 0$.
The differential equation $(y')^2+x^2=0$ has no solution because $(y')^2$ and $x^2$ are both non-negative.

A relation $f(x,y)=0$ is called an implicit solution of the differential equation (v) on an interval $I$ if

there exists a function $y=\phi(x)$ such that $f\left[x,\phi(x)\right]=0$ for every $x$ in the interval $I$
$\phi(x)$ satisfies Equation (v); that is, \[F\left(x,\phi(x), \phi'(x),\cdots, \phi^{(n)}(x)\right)=0 \quad \text{for every } x \in I\]

For example, consider the following differential equation:

\[\frac{dy}{dx}=\frac{x+y+1}{y^2-x-3}.\]

We can show that

\[3x^2+6xy+6x-2y^3+18y-4=0\]

is an implicit solution of this differential equation.

Example: Show that $3x^2+6xy+6x-2y^3+18y-4=0$ is the implicit solution of $\dfrac{dy}{dx}=\dfrac{x+y+1}{y^2-x-3}$.

Solution

Let's use implicit differentiation to find $\dfrac{dy}{dx}$: \[\frac{d}{dx}\left[3x^2+6xy+6x-2y^3+18y\right]-\frac{d}{dx}4=0\]\[6x+\underbrace{6y+6x\frac{dy}{dx}}_{\frac{d}{dx}(6xy)}+6-\underbrace{2\times 3 y^2\frac{dy}{dx}}_{\frac{d}{dx}(2y^3)}+18\frac{dy}{dx}=0\]\[\left(6x-6y^2+18\right)\frac{dy}{dx}+6(x+y+1)=0\]\[\frac{dy}{dx}=-\frac{x+y+1}{x-y^2+3}=\frac{x+y+1}{y^2-x-3}\]

General, Particular, and Singular Solutions

Solutions to differential equations can be classified into three types: general, particular, and singular.

An ordinary differential equation of order $n$ generally has a solution containing $n$ arbitrary constants. We call such a solution the general solution.

If the equation is linear, the general solution contains exactly n arbitrary parameters, and every solution of the equation is obtained by assigning specific values to these parameters.

For example,

\[y=c_1 \sin x+ c_2\cos x\]

where $c_1$ and $c_2$ are arbitrary constants, is the general solution of the equation $y^{\prime\prime}+y=0$.

A particular solution is a solution obtained from the general solution by assigning specific values to its constants.

A singular solution is a solution without arbitrary constants that cannot be derived from the general solution by assigning specific values to its constants. Such solutions arise only in nonlinear differential equations.

Let's consider the following differential equation: $ \left(\frac{dy}{dx}\right)^2-4y=0. $ Notice that this equation is nonlinear because $dy/dx$ is raised to the power of two, rather than one. We can rewrite the equation as $ \frac{dy}{dx}=\pm 2\sqrt{y} $ If $y\neq 0$, we can divide both sides by $2\sqrt{y}$ and get $ \frac{1}{2\sqrt{y}}\frac{dy}{dx}=\pm 1 $ which can also be written as $ \frac{dy}{2\sqrt{y}}=\pm dx $ Integrating both sides, we get $ \sqrt{y}=\pm(x+c) $ or $ y=(x+c)^2 $ Because it contains the arbitrary constant $c$, $y=(x+c)^2$ is the general solution of the given equation. It represents a family of parabolas opening upward, shifted left or right depending on the value of $c$.

Particular solutions can be obtained by assigning certain values to $c$. For example,

$ y=(x+1)^2 $

is a particular solution.

In the solution process, we assumed that $y\neq 0$. This assumption allowed us to divide both sides by $2\sqrt{y}$. If we substitute $y=0$ into the original equation, we find that $y=0$ is also a solution.

Since we cannot get $y=0$ from the general solution $y=(x+c)^2$ by choosing a value for $c$, $y=0$ is called a singular solution.