Module 5 First Order Differential Equations

Module S First Order Differential Equations

Differential Equations (DE)

• A differential equation contains derivatives of an unknown function.
• Examples:
  • $\frac{d^2y}{dx^2} + \frac{d^3y}{dx^3} = 2$
  • $\frac{dy}{dx} = xy + 2x$
  • $\frac{dy}{dx} + 2y + 3 = x$
• Algebraic equations examples:
  • $x + y = 2$
  • $x^2 + xy = y^2$
• Rectilinear motion example: acceleration $a = \frac{dv}{dt} = -10$, initial velocity $u = 10$

Order of a DE

• The order of a DE refers to the highest derivative in the equation.
• Examples:
  • $y' = xy + 2x$ (1st order DE)
  • $y'' + 2y + 3 = x$ (2nd order DE)
  • $y''' = 2y' - 3$ (3rd order DE)
• This module focuses only on 1st order DEs.

Direction Fields

• Direction fields can be used to understand and visualize the behavior of solutions to a DE.
• Consider 1st order DEs of the form $\frac{dy}{dx} = f(x,y)$.
• Example: Consider the DE $\frac{dy}{dx} = x + y - 1$ (i.e., $f(x,y) = x + y - 1$).
• The DE says that the slope of the solution $y(x)$ at $(x,y)$ is $x + y - 1$.
• A direction field is generated by plotting slope lines at an array of points.
• Example points:
  • At $(x,y) = (0,0)$, $\frac{dy}{dx} = 0 + 0 - 1 = -1$
  • At $(x,y) = (1,1)$, $\frac{dy}{dx} = 1 + 1 - 1 = 1$
  • At $(x,y) = (2,-1)$, $\frac{dy}{dx} = 2 + (-1) - 1 = 0$

Matching DEs with Direction Fields

• Example: Match each of the DEs with its direction field.
  • I. $y' = y^2 + y - 2$
  • II. $y' = y - 2x$
  • III. $y' = y^2 - y - 2$
  • IV. $y' = -y^2 - y + 2$
  • V. $y' = 1 - xy$
  • VI. $y' = y + xy$
• Solutions:
  • I matches with E.
  • II matches with A.
  • III matches with B.
  • IV matches with C.
  • V matches with D.
  • VI matches with F.

Plotting Solutions on a Direction Field

• To plot a solution on a direction field, we must specify a starting point $(x<em>0, y</em>0)$.
• Then, move from that starting point in a direction parallel to the surrounding field lines.
• Important: You cannot cross a slope line if you are moving parallel to it.

Initial Value Problem (IVP)

• A DE with a starting point is called an initial value problem (IVP).
• $\frac{dy}{dx} = f(x,y), \quad y(x<em>0) = y</em>0$
  • $\frac{dy}{dx} = f(x,y)$ is the DE.
  • $y(x<em>0) = y</em>0$ is the starting point/initial value.
• Example: Plot the solutions to the following IVPs using the provided direction field for $\frac{dy}{dx} = x + y - 1$.
  • a) $\frac{dy}{dx} = x + y - 1, \quad y(0) = 1$
  • b) $\frac{dy}{dx} = x + y - 1, \quad y(0) = 0$
  • c) $\frac{dy}{dx} = x + y - 1, \quad y(-1) = 0$

Autonomous DES

• In previous sections, we considered direction fields of 1st order DEs of the form $\frac{dy}{dx} = f(x,y)$.
• A 1st order DE is called autonomous if it has the form $\frac{dy}{dx} = f(y)$.
  • i.e., The RHS depends only on the dependent variable $y$, not the independent variable $x$.
• Examples of the DEs considered previously:
  • $\frac{dy}{dx} = y^2 + y - 2$
  • $\frac{dy}{dx} = y^2 - y + 2$
  • $\frac{dy}{dx} = -y^2 - y + 2$
  • Autonomous DES
  • $\frac{dy}{dx} = x + y - 1$
  • $\frac{dy}{dx} = y - 2x$
  • $\frac{dy}{dx} = 1 - xy$
  • $\frac{dy}{dx} = y + xy$
  • Not autonomous DES
• If we compare direction fields, we see that the slopes of an autonomous DE do not depend horizontally on $x$.
  • That is, the slope does not change as we move horizontally on the direction field.

Equilibrium Solutions of Autonomous DES

• An equilibrium solution of an autonomous DE $\frac{dy}{dx} = f(y)$ is a solution of the form $y = constant$.
• Note: If $y = constant$, then $\frac{dy}{dx} = 0$. So, we find equilibrium solutions by solving $f(y) = 0$.
• Example: Determine the equilibrium solutions of $\frac{dy}{dx} = y^2 + y - 2$.
  • Equilibrium sol $\Rightarrow \frac{dy}{dx} = 0$
  • $\Rightarrow 0 = y^2 + y - 2$
  • $\Rightarrow 0 = (y + 2)(y - 1)$
  • $\Rightarrow y = -2, \quad y = 1$

More Examples of Equilibrium Solutions

• Determine the equilibrium solutions of $\frac{dy}{dx} = y^3 - 4y^2 + 4y$.
  • Equilibrium sol $\Rightarrow \frac{dy}{dx} = 0$
  • $\Rightarrow 0 = y^3 - 4y^2 + 4y$
  • $\Rightarrow 0 = y(y^2 - 4y + 4)$
  • $\Rightarrow 0 = y(y - 2)^2$
  • $\Rightarrow y = 0, \quad y = 2$
• Determine the equilibrium solutions of:
  • a) $\frac{dy}{dx} = -y^2 + y + 6$
    • Need $\frac{dy}{dx} = 0$
    • $\Rightarrow 0 = -y^2 + y + 6$
    • $\Rightarrow 0 = -(y^2 - y - 6)$
    • $\Rightarrow 0 = -(y - 3)(y + 2)$
    • $\Rightarrow y = -2, \quad y = 3$
  • b) $\frac{dy}{dx} = y \cos(y)$
    • Need $\frac{dy}{dx} = 0$
    • $\Rightarrow 0 = y \cos(y)$
    • $\Rightarrow y = 0 \quad or \quad \cos(y) = 0$
    • $\Rightarrow y = 0, \quad y = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, …$
    • $y = (2n+1)\frac{\pi}{2}, \quad n \in \mathbb{Z}$

Stability of Equilibrium Solutions

• An equilibrium solution $y = c$ is:
  • Stable if all nearby solutions approach $c$ as the independent variable $(x) \rightarrow \infty$.
  • Unstable if all nearby solutions move away from $c$ as the independent variable $(x) \rightarrow \infty$.
• Can determine stability using a sign diagram.
• Example: Determine the stability of the equilibrium solutions of $\frac{dy}{dx} = y^2 + y - 2$.
  • Previously found $y = 1$ and $y = -2$ are equilibrium solutions.
  • Test points:
    • y = 2 \Rightarrow \frac{dy}{dx}(2) = 2^2 + 2 - 2 = 4 > 0
    • y = 0 \Rightarrow \frac{dy}{dx}(0) = 0^2 + 0 - 2 = -2 < 0
    • y = -3 \Rightarrow \frac{dy}{dx}(-3) = (-3)^2 - 3 - 2 = 4 > 0
  • Hence, $y = 1$ is an unstable sol, and $y = -2$ is a stable sol.

More Examples of Stability

• Determine the equilibrium solutions of the following DE and determine the stability of these solutions:
  • $\frac{dy}{dx} = y^3 - 4y^2 + 4y$
    • Previously found that $y = 0$ and $y = 2$ are equilibrium solutions.
    • Test points:
      • y = 3 \Rightarrow \frac{dy}{dx}(3) = 3^3 - 4(3)^2 + 4(3) = 3 > 0
      • y = 1 \Rightarrow \frac{dy}{dx}(1) = 1^3 - 4(1)^2 + 4(1) = 1 > 0
      • y = -1 \Rightarrow \frac{dy}{dx}(-1) = (-1)^3 - 4(-1)^2 + 4(-1) = -9 < 0
    • So, $y = 0$ is unstable, and $y = 2$ is semi-stable (stable from below, unstable from above).
• Determine the equilibrium solutions of the following DEs & their stability.
  • a) $\frac{dy}{dx} = -y^2 + y + 6$
    • From previous example, $y = -2$ and $y = 3$ are equilibrium solutions.

Applications Examples

• Applications
• A. Natural growth
  • \frac{dP}{dt} = rP
    • P = population Size
    • t = time
    • r = Growth rate
  • r=1
    • Equilibrium sol
• \frac{dP}{dt} = P and P(0) = 5

Separable DES

• A 1st order DE is called separable if it can be written in the form
  • \frac{dy}{dx} = f(y) g(x)
  • or
  • all y stuff = all x stuff
• dy/(f(y)) = g(x)dx
• Examples
  • Separable DE
    • \frac{dy}{dx} = cos(x)y,
    • g(x) = cos(x)
    • f(y) = y
    • \frac{dy}{dx} = \frac{x}{y}
    • g(x) = x
    • f(y) = y
  • Not separable
    • \frac{dy}{dx} = cos(x)+ y^2
• Solution technique
• \frac{dy}{dx} = dy = f(y)g(x)
• Integrate both sides
• \int\frac{dy}{f(y)} = \int^{} g(x) dx
• Remember differential and implicit form
• Differential form example
  • General solution to the DE \frac{dy}{dx} = xy
• Separate the dy and dx
• \frac{dy}{y} = xdx
• note: integral dy/f(y) = \int f' divide dx
• \int^{}\frac{dy}{y} = \int x dx
• Lnly = \frac{X^2}{2} + C
• implicit solution
• Index laws to solve
• y = e^((\frac{x^2}{2})+c)
• y = ae^\frac{x^2}{2}
• explicit general solution
• y = Ae^(x^2/2)
• This is called the general solution to the DE

Particular solutions and the IVP

• We can get a particular solution by specifiying an initial condition (IC).
  • This allows us to determine the particular value of the constant A.
  • Eg. Solve the IVP \frac{dy}{dx} = xy and y(0) = 3.
    • Solution:
      • From the previous example, y = A e^(x^2/2).
      • Applying IC: 3 = A e^(0) = A. So, A = 3.

More Separable DE Solutions

• Determine the general solution to the DE \frac{dy}{dx} = xy^2.
  • Then, solve the IUP \frac{dy}{dx} = xy^2, y(1)=2
  • Separable DE
    • \frac{dy}{y^2} = xdx
• Integrate them
  • \int y^-2 dy = \int xdx
  • -y^-1 = \frac{x^2}{2} + C
  • y = \frac{-1}{x^2/2 + C}

Cosine IVP and Separable General Solution Examples

• Solve the TVP \frac{dy}{dx} = \frac{cos(x)}{e^y} and y(0) = 0
• Determine the general sol \frac{dy}{dx} = \frac{y}{x}
• Solved this equation for Separable DE
  • \int e^y = \int cos(x)dx
  • e^(y) = sin(x) + C
  • Take the Ln of each side
    • Lne^y = ln(sin(x) + C)
  • y = ln(sin(x) + C)
• Applying the initial conditions
  • y = ae^\frac{x^2}{2}
• Applying the initial conditions
  • y(0) =3
  • y=3 Ae^(x^2/2)

Autonomous DE and More

• Autonomous DE and more
• \frac{dy}{dx}= \frac{dy}{(y^2 +y - 2)} = dx
• (y+2)(y-1)=0
  • Equilibrium sols are \frac{dy}{(y+2)} and and (y-1)=0
• need A/y+2 + B/ y-1
• If y=1 then
  • \frac{A}{(y-1)} + B/3 = 1/ y+2)(y-1)

Application of natural Growth Formulas

• Separable
• \frac{dp}{p} = dt
  • Inp = t + c OR p = A e^( t/2)
• Apply the variables
  • t=5 then evaluate a long answer
• Logistic growth: grow according to the logistic equation
  • \frac{dp}{dt} = r * - \frac{p}{k}
  • K = carrying capacity
• If r = (1/ 2), k= 100 P growings according y(o)=50
  • Separable solution for P (t)

Logistic Growth Explained

• Logistical Growth problems
• A new substance that has a reactions in two froms $P * Q \rightarrow X$
  • Q >=x, or the concentration of the products
• What concentrations in x from this

Liner DES

• A 1st order DE is called linear if it can be written in the form

• \frac{dy}{dx} + P(x)y = Q(x).

• Eg.

• \frac{dy}{dx} + (cos(x)) y = e^2

• Where p(x) = cos(x)

• Examine y term

• Solution technique

• step 1: calculate Integrationg facor

• 1 = e^( integralP(xx)dx)

• \int Ixy Q(x)dx

• Combine Ixy etc info the general solution problem

Doses the integrate really work?

• Determine the general solution to the DE.

• \frac{dy}{dx} + 3y + 4

• Solution for form of the DE

  • Step 1: calculate integrating factor
  • Ix = \int e^(p(x). dx) = \int e^(3 dx)

ICs and IUPs

• To determine a value for 'C', we need to specify an IC to give an IUP.
• Eg. Solve the IUP $\frac{dy}{dx} + 3y = 4, \quad y(0) = 1$.
  • Solution:
    • From previous example, $y = \frac{4}{3} + Ce^{-3x}$.
    • $y(0) = 1 \Rightarrow 1 = \frac{4}{3} + Ce^{-3(0)} \Rightarrow 1 = \frac{4}{3} + C \Rightarrow C = -\frac{1}{3}$.

More liner IVP and solutions

• linear

• \int1dx

• = \int e ^ -1dx = e∧ -x

• Step 2:

• IxyQ(x) dx = \inte∧xy(x)dx

• Let u=x

• dv = \int e^( -x)

• Then by parts -\int vdu

General solutions and formulas for solving

• General solutions formulas
  • Solve the following IUPs x \frac{dy}{dx} + 2y = 3
    • Obtain the general solution.
    • Solve: Liner convert to standard form
    • \frac{dy}{dx} + \frac{{3}}{x}y = 3
• 2) Sp(x)/dx = \int( dx/x)2=e^(2lnx)
• )simplify y (x), See previous example

Applications: Linear Model and Equations.

• Models with liner equations and forumls

• An Object of mass m is falling under gravity

• \frac{dv}{dt} mg+ km= 100-2v/10

  • V(0 = 0/ liner and separable