A differential equation (DE) is an equation containing the derivatives of one or more dependent variables with respect to one or more independent variables
Notation
Leibniz notation \(~\displaystyle\color{blue}{\frac{d^2y}{dx^2}}\)
Prime notation \(~\color{blue}{y''}\)
Newton’s dot notation \(~\color{blue}{\ddot{y}}\)
Subscript notation \(~\color{blue}{u_{xx} + u_{yy} = 0}\)
Ordinary differential equation (ODE): \(\,\) Derivatives are with respect to a single independent variable
\[\frac{dx}{d\color{blue}{t}} + \frac{dy}{d\color{blue}{t}} = 3x + 2y\]
Partial differential equation (PDE): \(\,\) Derivatives are with respect to two or more independent variables
\[\frac{\partial u}{\partial \color{blue}{y}} = -\frac{\partial u}{\partial \color{blue}{x}}\]
The order of an ODE or PDE is the order of the highest derivative in the equation
\[\underbrace{\color{blue}{\frac{d^2y}{dx^2}}}_{\mathrm{highest\; order}} +5\left ( \frac{dy}{dx} \right )^3 - 4y = e^x\]
\[\underbrace{2\color{blue}{\frac{\partial^4 u}{\partial x^4}}}_{\mathrm{highest\; order}} +\frac{\partial^2 u}{\partial t^2} = 0\]
An \(n\)-th order ODE, \(F\left( x, y, y', \cdots, y^{(n)} \right) = 0\), \(\,\)is linear in the variable \(y\,\) if \(\,F\) is linear in \(y, y',\cdots,y^{(n)}\)
An ODE is nonlinear if
Similar to integration, we obtain a solution to a first-order differential equation containing an arbitrary constant \(c\)
A solution with a constant \(c\) represents a set \(G(x,y,c)=0\) of solutions, called a one-parameter family of solutions
An \(n\)-parameter family of solutions \[ G(x,y,c_1,c_2,\cdots,c_n)=0 \]
solves an \(n\)-th order differential equation
Two or more equations involving the derivatives of two or more unknown functions of a single independent variable are called a system of differential equations
\[ \begin{aligned} \frac{dx}{dt} &= f(t,x,y) \\ \frac{dy}{dt} &= g(t,x,y) \end{aligned}\]
A solution of a system is a set of functions defined on a common interval \(I\) that satisfies each equation of the system on this interval
In an IVP, we seek a solution \(y(x)\) of a differential equation so that \(y(x)\) satisfies initial conditions at \(x_0\)
\(n\)-th order differential equation and Initial conditions
\[ \begin{aligned} \frac{d^ny}{dx^n} &= f \left ( x,y,y',\cdots,y^{(n-1)} \right ) \\ y(x_0) &= y_0, \;y'(x_0) = y_1, \cdots, \;y^{(n-1)}(x_0) = y_{n-1} \\ \\ &\Downarrow \\ \\ \frac{d\mathbf{y}}{dx} &= \mathbf{f}(x, \mathbf{y}), \;\;\mathbf{y}(x_0)=\mathbf{y}_0 \end{aligned}\]
When solving an IVP, consider whether a solution exists and whether the solution is unique
Theorem 1.1 gives conditions that are sufficient to guarantee the existence and uniqueness of a solution \(y(x)\) to a first-order IVP
Let \(R\) be a rectangular region in the \(xy\)-plane defined by \(a\le x\le b\), \(c\le y\le d\) that contains the point \((x_0,y_0)\) in its interior. If \(f(x,y)\) and \(\partial f/\partial y\) are continuous on \(R\), then there exists some interval \(I_0\): \((x_0-h,\,x_0+h)\), \(h>0\), contained in \([a,b]\), and a unique function \(y(x)\), defined on \(I_0\), that is a solution of the IVP \(\;dy/dx=f(x,y)\), \(y(x_0)=y_0\)
Example \(\,\) First-order IVP
\(y(x)=ce^x\) is a one-parameter family of solutions of the first-order DE \(~y'=y\)
Find a solution of the first-order IVP with initial condition \(y(0)=3\)
Solution
From the initial condition, we obtain \(3=ce^0\)
Solving, we find \(c=3\)
The solution of the IVP is \(y=3e^x\)
Example \(\,\) Second-order IVP
\(x(t) = c_1 \cos 4t +c_2 \sin 4t\) \(\,\)is a two-parameter family of solutions of the second-order DE \(~x''+16x=0\)
Find a solution of the second-order IVP with initial conditions \(~x\left( \frac{\pi}{2} \right) =-2, \;x'\left (\frac{\pi}{2} \right)=1\)
Solution
Example \(\,\) Radioactive Decay
In modeling the phenomenon, it is assumed that the rate \(\displaystyle\frac{dA}{dt}\) at which the nuclei of a substance decay is proportional to the amount \(A(t)\) remaining at time \(t\) given an initial amount of radioactive substance on hand \(A_0\)
\[\frac{dA}{dt} = -kA, \;A(0)=A_0\]
This differential equation also describes a first-order chemical reaction
Example \(\,\) Draining Tank
Torricelli’s law states that the exit speed \(v\) of water through a sharp-edged hole at the bottom of a tank filled to a depth \(h\) is the same as the speed it would acquire falling from a height \(h\), \(\,\)\(v=\sqrt{2gh}\)
The volume \(V\) of water leaving the tank per second is the area of the hole \(A_h\) times the exit speed \[\frac{dV}{dt}=-A_h\sqrt{2gh}\]
The volume in the tank at \(t\) \(\,\)is\(\,\) \(V(t)=A_w h\), \(\,\) in which \(\,\) \(A_w\) is the constant area of the upper water surface
Combining expressions gives the differential equation for the height of water at time \(t\) \[\frac{dh}{dt}=-\frac{A_h}{A_w}\sqrt{2gh}\]
| Concept | Key point |
|---|---|
| Order | order of the highest derivative in the equation |
| Linear ODE | \(a_n(x)y^{(n)}+\cdots+a_1(x)y'+a_0(x)y=g(x)\): \(\,\)linear in \(y, y', \cdots, y^{(n)}\) |
| Solution | a function \(\phi\) on an interval \(I\) that reduces the ODE to an identity; \(\,\)explicit \(y=\phi(x)\) or implicit \(G(x,y)=0\) |
| \(n\)-parameter family | \(G(x,y,c_1,\cdots,c_n)=0\) solves an \(n\)-th order ODE |
| Initial-value problem | ODE with \(y(x_0)=y_0,\ y'(x_0)=y_1,\ \cdots,\ y^{(n-1)}(x_0)=y_{n-1}\) |
| Existence and uniqueness | \(y'=f(x,y)\), \(\,y(x_0)=y_0\): \(\,\)if \(f\) and \(\partial f/\partial y\) are continuous on a rectangle containing \((x_0,y_0)\), \(\,\)there is exactly one solution near \(x_0\) |
| Models | radioactive decay \(\dfrac{dA}{dt}=-kA\); \(\quad\) draining tank \(\dfrac{dh}{dt}=-\dfrac{A_h}{A_w}\sqrt{2gh}\) |
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1.2: 1. \(~y=1/(1 +c_1 e^{-x})\) is a one-parameter family of solutions of the first-order DE \(y'=y -y^2\). \(\,\) Find a solution of the first-order IVP consisting of this differential equation and the given initial condition:
\[y(0)=-\frac{1}{3}\]
Solution
\[ \begin{aligned} y &= \frac{1}{1 +c_1 e^{-x}}\\ &\Downarrow {\scriptstyle \text{at } x = 0}\\ -\frac{1}{3} &= \frac{1}{1 +c_1} \\ &\Downarrow \\ c_1 &= -4 \\ &\Downarrow \\ y &= \frac{1}{1 -4 e^{-x}} \end{aligned}\]
1.2: 3. \(~y=1/(x^2 +c)\) is a one-parameter family of solutions of the first-order DE \(y' +2xy^2=0\). Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval \(I\) over which the solution is defined
\[y(2)=\frac{1}{3}\]
Solution
\[ \begin{aligned} y &= \frac{1}{x^2 +c}\\ &\Downarrow {\scriptstyle \text{at } x = 2}\\ \frac{1}{3} &= \frac{1}{4 +c} \\ &\Downarrow \\ c = -1 \, &, \;\; I =(1,\infty) \\ &\Downarrow \\ y &= \frac{1}{x^2 -1} \end{aligned}\]
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-5, 5, 200)
y = 1.0 /(x**2 -1)
threshold = 20
y[y > threshold] = np.inf
y[y <-threshold] = np.inf
plt.plot(x,y)
plt.scatter(2, 1/3, color='red')
plt.title(r'$y=\frac{1}{x^2 -1}$')
plt.xlabel('$x$')
plt.ylabel('$y$')
plt.grid()
plt.show()
1.2: 7. \(\,x(t)=c_1\cos t +c_2\sin t\) is a two-parameter family of solutions of the second-order DE \(x''+x=0\). Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions
\[x(0)=-1, \; x'(0)=8\]
Solution
\[ \begin{aligned} x(t) &= c_1\cos t +c_2\sin t \\ &\Downarrow \;{\scriptstyle x(0)=-1, \; x'(0)=8}\\ -1 &= c_1 \\ 8 &= c_2 \\ &\Downarrow \\ x(t) &= -\cos t +8\sin t \end{aligned}\]
1.2: 17. \(\,\) Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point \((x_0, y_0)\) in the region.
\[\frac{dy}{dx}=y^{2/3}\]
Solution
Step 1: \(~\)Identify the Function and Its Partial Derivative
Let:
\[f(x, y) = y^{2/3}\]
Then the partial derivative with respect to \(y\) is:
\[\frac{\partial f}{\partial y} = \frac{2}{3} y^{-1/3}\]
Step 2: \(~\)Analyze Continuity
\(f(x, y) = y^{2/3}\) is continuous for all real values of \(x\) and \(y\)
However, \(\frac{\partial f}{\partial y} = \frac{2}{3} y^{-1/3}\) is undefined at \(y = 0\), since \(y^{-1/3}\) is not defined at \(y = 0\)
Step 3: \(~\)Apply the Existence and Uniqueness Theorem
\(~\)
1.2: 29.
By inspection, find a one-parameter family of solutions of the differential equation \(xy'=y\). Verify that each member of the family is a solution of the initial value problem \(xy'=y, \,y(0)=0\)
Explain part 1. by determining a region \(R\) in the \(xy\)-plane for which the differential equation \(xy'=y\) would have a unique solution through a point \((x_0, y_0)\) in \(R\)
Verify that the piecewise-defined function
\[y= \begin{cases} 0, & x < 0 \\ x, & x \ge 0 \end{cases}\]
satisfies the condition \(y(0)=0\). Determine whether this function is also a solution of the initial-value problem in part 1
Solution
By inspection, \(\,y=cx\) satisfies \(xy'=x\cdot c=y\) for every constant \(c\). \(\,\)Every member satisfies \(y(0)=c\cdot 0=0\), \(\,\)so the IVP \(xy'=y\), \(y(0)=0\) has infinitely many solutions
In normal form \(\,y'=f(x,y)=y/x\), \(\,\)and \(f\) and \(\partial f/\partial y=1/x\) are continuous wherever \(x\neq 0\). \(\,\)So a region \(R\) is either half-plane \(x>0\) or \(x<0\); \(\,\)through any point \((x_0,y_0)\) with \(x_0\neq 0\) there is a unique solution. \(\,\)The initial point \((0,0)\) lies in no such region, \(\,\)so the theorem does not apply there, \(\,\)which is consistent with part 1
The piecewise function satisfies \(y(0)=0\), \(\,\)and it satisfies \(xy'=y\) for \(x<0\) (\(0=0\)) and for \(x>0\) (\(x\cdot 1=x\)). \(\,\)However, \(\,\)its left derivative at \(x=0\) is \(0\) and its right derivative is \(1\), \(\,\)so \(y'(0)\) does not exist. \(\,\)Therefore it is not a solution of the IVP on any interval containing \(x=0\)
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1.3: 5. \(\,\) A cup of coffee cools according to Newton’s law of cooling. Use data from the graph of the temperature \(T(t)\) in figure to estimate the constants \(T_m\), \(T_0\), and \(k\) in a model of the form of the first-order initial-value problem
\[ \frac{dT}{dt}=k(T -T_m), \;\;T(0)=T_0 \]
Solution
The solution of the IVP is \(\,T(t)=T_m+(T_0-T_m)e^{kt}\)
\(T_0=T(0)\) is read at \(t=0\): \(\,T_0\approx 175\)
\(T_m=\lim_{t\to\infty}T(t)\) is the level the curve approaches: \(\,T_m\approx 75\)
\(k\) follows from one more point on the curve. \(\,\)The curve reaches \(T\approx 100\) at \(t\approx 17\) min, \(\,\)so
\[ k=\frac{1}{t}\ln\frac{T-T_m}{T_0-T_m}\approx \frac{1}{17}\ln\frac{100-75}{175-75}=\frac{\ln 0.25}{17}\approx -0.08\ \text{min}^{-1} \]
(equivalently, \(\,k\) is the slope of \(\ln\left[(T-T_m)/(T_0-T_m)\right]\) plotted against \(t\))
1.3: 19. After a mass \(m\) is attached to a spring, it stretches \(s\) units and then hangs at rest in the equilibrium position as shown in Figure (b). After the spring/mass system has been set in motion, let \(x(t)\) denote the directed distance of the mass beyond the equilibrium position. As indicated in Figure (c), assume that the downward direction is positive, that the motion takes place in a vertical straight line through the center of gravity of the mass, and that the only forces acting on the system are the weight of the mass and the restoring force of the stretched spring. Use Hooke’s law: The restoring force of a spring is proportional to its total elongation. Determine a differential equation for the displacement \(x(t)\) at time \(t\)
Solution
Equilibrium (Figure (b)): \(\,\)the spring is stretched by \(s\) and the weight is balanced by the spring force (Hooke’s law, spring constant \(k\))
\[ mg=ks \]
In motion (Figure (c)): \(\,\)the total elongation is \(s+x\) (downward positive), \(\,\)so the forces on the mass are the weight \(mg\) (down) and the restoring force \(-k(s+x)\). \(\,\)Newton’s second law gives
\[ m\frac{d^2x}{dt^2}=mg-k(s+x)=\underbrace{mg-ks}_{0}-kx \]
Hence
\[ m\frac{d^2x}{dt^2}+kx=0 \]
Measuring \(x\) from the equilibrium position removes the weight from the equation