# Preface

This section presents some basic properties of second and higher order differential equations.

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# Differential Equations of higher order

Recall from calculus that derivatives of a smooth function y(x) are denoted as $$y'(x)$$ (Lagrange notation) or $${\text d}y /{\text d}x$$ (Leibniz notation), or, in case of time variable t, as $$\dot{y} .$$ Second derivatives are denoted by $$y''(x)$$ or $${\text d}^2 y /{\text d}x^2$$ or $$\ddot{y} ,$$ respectively. Following Euler, we will denote by $$\texttt{D} = {\text d}/{\text d}x$$ the (unbounded) derivative operator. Usually, the notations x or t stand for the independent variables and will be widely used. Higher order derivatives have similar notation. For instance, $$y^{(4)} (x)$$ stands for the fourth derivative of function y(x).

A second order differential equation in the normal form is as follows: $$\label{E41.1} \frac{{\text d}^2 y}{{\text d} x^2} = F\left( x,y,\frac{{\text d}y}{{\text d}x} \right) \qquad \mbox{or} \qquad y'' = F\left( x,y,y' \right) ,$$ where F(x, y, p) is some given function of three variables. If the function F(x,y,p) is linear in variables y and p (that is, $$F(x,ay_1 + by_2 , p) = a\,F(x,y_1 , p) + b\,F(x, y_2 , p)$$ for any constants a, b, and similar for variable p), then $$y'' = F\left( x,y, y' \right)$$ is called linear.

For example, the equation $$y'' = \sin x + 3 y^2 + 2 \left( y' \right)^2$$ is a second order nonlinear differential equation, while the equation $$y'' = (\cos x)\,y$$ is a linear one.

A function $$y=\phi (x)$$ is a solution of $$y'' = F\left( x,y, y' \right)$$ in some open interval $$a < x < b$$ (perhaps infinite), having derivatives up to the second order throughout the interval, if φ(x) satisfies the differential equation in an open interval (𝑎, b), that is,

$\frac{{\text d}^2 \phi}{{\text d} x^2} = F\left( x,\phi (x),\frac{{\text d}\phi}{{\text d}x} \right) \qquad \mbox{ for all } \ x\in (a,b).$

For many of the differential equations to be considered, it will be found that solutions of $$y'' = F\left( x,y, y' \right)$$ can be included in one formula, either explicit

$y=\phi (x, C_1 , C_2 ) \qquad \mbox{or implicit} \qquad \Phi (x,y, C_1 , C_2 )=0,$
where C1 and C2 are arbitrary constants. Such a solution is referred to as the general solution of the differential equation of the second order in either explicit or implicit form. Choosing specific values of the constants C1 and C2, we obtain a particular solution of $$y'' = F\left( x,y, y' \right) .$$ All solutions can be so found except, possibly, singular and/or equilibrium solutions.

Second order differential equations are widely used in science and engineering to model real world problems. The most famous second order differential equation is Newton's second law of motion, $$m\,\ddot{y} = F\left( t, y, \dot{y} \right) ,$$ which describes a one-dimensional motion of a particle of mass m moving under the influence of a force F. In this equation, y=y(t) is the position of a particle at a time t, $$\dot{y} = {\text d}y/{\text d}t$$ is its velocity, $$\ddot{y} = {\text d}^2 y/{\text d}t^2$$ is its acceleration, and F is the total force acting on the particle.

For given two numbers y0 and y1, we impose the initial conditions on y(x) in the form

$$\label{E41.2} y( x_0 )=y_0, \quad y' ( x_0 ) = y_1 .$$
The differential equation $$y'' = F\left( x,y, y' \right)$$ with the initial conditions $$y( x_0 )=y_0, \quad y' ( x_0 ) = y_1$$ is called the initial value problem (IVP for short).

Theorem 1: Suppose that F, $$\partial F/\partial y,$$ and $$\partial F/\partial y'$$ are continuous in a closed 3-dimensional domain Ω in xyy'-space, and the point $$(x_0 ,y_0 , y'_0 )$$ belongs to Ω. Then the initial value problem $$y'' = F\left( x,y, y' \right) , \quad y( x_0 )=y_0, \quad y' ( x_0 ) = y_1$$ has a unique solution $$y=\phi (x)$$ on an x-interval in Ω containing x0.

Since the differential operator $$\texttt{D} = {\text d}/{\text d}x$$ is an unbounded operator, it is convenient to rewrite the initial value problem in equivalent integral form:
$$\label{EqODE2.3} y(x) = y( x_0 ) + y' \left( x_0 \right) \left( x - x_0 \right) + \int_{x_0}^x \left( x-t \right) F\left( t, y(t) , y' (t) \right) {\text d} t .$$

The general linear differential equation of the second order is an equation that can be written as
$$\label{E41.3} a_2 (x) \frac{{\text d}^2 y}{{\text d} x^2} + a_{1} (x) \frac{{\text d}y}{{\text d}x} + a_{0} (x) y(x) = g(x).$$
The functions $$a_0 (x), a_1 (x), a_2 (x),$$ are referred to as coefficients of the differential equation and g(x) is a given function, known as driving term, forcing term, or nonhomogeneous term; they all are assumed to be independent of dependent variable y. We will denote by x and/or t the independent variables. When a differential equation cannot be written in the above form, then it is called nonlinear. If the function $$a_2 (x)$$ has no zeroes on some interval, then we can divide both sides of the equation by $$a_2 (x)$$ to obtain its normalized form:
$$\label{E41.5} y'' (x) + p (x) \,y' (x) + q (x)\, y(x) = f(x).$$

The points where the coefficients of $$y'' (x) + p (x) \,y' (x) + q (x)\, y(x) = f(x)$$ are discontinuous or undefined are called the singular points of the equation. These points are usually not used in the initial conditions except some cases. For example, the equation $$(x^2 -1)\,y'' + y=1$$ has two singular points x=1 and x=-1 that must be excluded. If in opposite, the initial condition $$y(1) =y_0$$ is imposed, then the differential equation dictates that $$y_0 =1 ;$$ otherwise, it has no solution.

Theorem 2: Let p(x), q(x), and f(x) be continuous functions on an open interval $$a < x < b.$$ Then, for each $$x_0 \in (a,b),$$ the initial value problem
$y'' + p(x)y' + q(x) y =f(x), \qquad y(x_0 ) =y_0 , \quad y' (x_0 ) = y_1$
has a unique solution for arbitrary specified real numbers y0, y1.

Theorem 3: A homogeneous differential equation $$y'' + p(x)\,y' + q(x)\, y = 0$$ upon substitution y(x) = u(x) v(x), where v is the square root of the Wronskian of the linearly independent solutions to the given differential equation,
$v' = - \frac{1}{2}\,p(x)\, v(x) \qquad \Longrightarrow \qquad v(x) = \exp \left\{ - \frac{1}{2} \int p(x)\,{\text d}x \right\} ,$
is reduced to the differential equation for u:
$u'' + \left( q(x) -\frac{1}{2}\, p' (x) - \frac{1}{4}\, p^2 \right) u = 0.$
Differentiating the function v(x), we get
\begin{align*} v' (x) &= v (x) \left( -\frac{1}{2}\, p (x) \right) , \\ v'' (x) &= v' (x) \left( -\frac{1}{2}\, p (x) \right) + v (x) \left( -\frac{1}{2}\, p (x) \right)' = v (x) \left( -\frac{1}{2}\, p (x) \right)^2 - \frac{1}{2}\, p' (x)\, v(x) \\ &= \left( \frac{1}{4} \,p^2 (x) - \frac{1}{2}\, p' (x) \right) v(x) . \end{align*}
Using the product rule
$y' = u'\,v + u\, v' \qquad\mbox{and} \qquad y'' = u''\,v + 2\,u'\,v' + u\, v'' ,$
we obtain
$y'' + p(x)\, y' + q(x)\,y = v\,u'' + \left( 2v' + p\,v \right) u' + \left( v'' + p\,v' + q\,v \right) u =0.$
Because we have a degree of freedom, we can choose v so that $$2\,v' + p\,v = 0.$$ Then the term containing the first derivative of u disappear, and we have the second order differential equation for u:
$u'' + \left( v'' + p\,v' + q\,v \right) u =0.$
Now we calculate the multiple of u:
$v'' + p\,v' + q\,v = \left( \frac{1}{4} \,p^2 (x) - \frac{1}{2}\, p' (x) + p(x)\left( -\frac{1}{2}\, p (x) \right) +q(x) \right) v(x) = \left( q(x) - \frac{1}{4} \,p^2 (x) - \frac{1}{2}\, p' (x) \right) v(x) .$
This leads to the required differential equation.    ▣

Equation $$y'' + p(x)y' + q(x) y =f(x)$$ is a particular case of the general linear differential equation of the n-th order

$$\label{E41.4} a_n (x) \frac{{\text d}^n y}{{\text d} x^n} + a_{n-1} (x) \frac{{\text d}^{n-1} y}{{\text d} x^{n-1}} + \cdots + a_0 (x) y(x) = g(x). %\eqno{(1.4)}$$
If g(x) is identically zero, the above equations are said to be homogeneous (the accent is on the syllable ge''); if g(x) is not identically zero, equations are called nonhomogeneous (or inhomogeneous or driven) and the function g(x) is referred to as the nonhomogeneous term, which is also called the input function or forcing function.

Theorem 4: Let functions $$a_0 (x), a_1 (x), \ldots , a_n (x)$$ and f(x) be defined and continuous on the closed interval $$a\leq x\leq b$$ with $$a_n (x) \neq 0$$ for $$x\in [a,b] .$$ Let x0 be such that $$a\leq x_0 \leq b$$ and let y0, y'0, $$\ldots ,\ y^{(n-1)}_0$$ be any constant. Then in the closed interval [a,b], there exists a unique solution y(x) satisfying the initial value problem:
$a_n (x) y^{(n)} + a_{n-1} (x) y^{(n-1)} + \cdots + a_{1} (x) y' + a_0 (x) y = g(x) ,$
$y( x_0 )= y_0 , \ y' ( x_0 ) = y'_0 , \ \ldots , \ y^{(n-1)} ( x_0 ) = y^{(n-1)}_0 .$
Example 1: Let us consider the initial value problem
$t(t^2 -4)\,y'' +t\,y' + (\ln |t|)\,y =0, \qquad y(1) =0, \quad y' (1) =2.$

Solution: To determine the validity interval (= the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution), we divide both sides of the differential equation by $$x(x^2 -4)=x(x-2)(x+2)$$ to obtain $$y'' + p(x)y' + q(x)y =f(x)$$ or

$y'' + \frac{1}{x(x-2)}\,y' + \frac{x}{(x-2)(x+2)}\,y = \frac{\sin x}{x} \cdot \frac{1}{(x-2)(x+2)} .$
The coefficient $$p (x) = 1/x(x-2)$$ is not defined at two singular points x=0 and x=2. Similarly, the functions $$q (x) = x/(x^2 -4)$$ and $$f(x) = \frac{\sin x}{x} \cdot\frac{1}{(x-2)(x+2)}$$ fail to be continuous at singular points $$x=\pm 2.$$ So we don't want to choose the initial point x0 as 0 or $$\pm 2.$$ For example, if x0 = 1, the given initial value problem has a unique solution in the open interval (0,2). If x0 = 3, the given initial value problem has a unique solution in the interval $$2 < x < \infty .$$ The behavior of a solution at these singularities requires additional analysis.

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Example 2:

Clear[x,y,t,z];
soln = DSolve[{x''[t] + 25 x[t] == 0, x[0] == 1, x'[0] == 0}, x[t], t]
Plot[x[t] /. soln, {t, -1, 2.5}]
s[t_] = x[t]/.soln[[1]]
Plot[s[t],{t,0,3},AxesLabel->{"t","Displacement"}]

soln = DSolve[{x''[t] + 17 x'[t] + 16 x[t] == 0, x[0] == 1, x'[0] == 5}, x[t], t];
s1[t_] = x[t] /. soln[[1]]
Out[32]= 1/5 E^(-16 t) (-2 + 7 E^(15 t))
Plot[s1[t], {t, 0, 3}, AxesLabel -> {"t", "Displacement"}]

>>>> displace1.jpg

When does the maximum excursion occur?

FindRoot[s1'[t] == 0, {t, 0}]
Out[9]= {t -> 0.101322}

What is the maximum excursion?
s1[t /. %]
Out[10]= 1.18603

soln = DSolve[{x''[t] + 2 x'[t] + 37 x[t] == 0, x[0] == -1, x'[0] == 5}, x[t], t]
Out[2]= -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])}}
s3[t_] = x[t] /. soln[[1]]

Clear[x,t];
soln = DSolve[{x''[t] + 2 x'[t] + 37 x[t] == 0, x[0] == -1,
x'[0] == 5}, x[t], t]
s3[t_] = x[t] /. soln[[1]]
Plot[s3[t], {t, 0, 3}, PlotRange -> {-1, 1}]

Out[19]= {{x[t] -> -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])}}
Out[20]= -(1/3) E^-t (3 Cos[6 t] - 2 Sin[6 t])
Out[21]=

>>>>> solution3.jpg
amplitude =
s3[t] /. c3_ Exp[c4_] (c1_ Cos[a_] + c2_ Sin[a_]) -> c3*Sqrt[c1^2 + c2^2]
Out[19]= -(Sqrt[13]/3)
Plot[{amplitude*Exp[-t], -amplitude*Exp[-t], s3[t]}, {t, 0, 3}, PlotStyle -> {Blue, Blue, Black}]

>>>>> amplitude.jpg

Forced Oscillations

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