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Wize University Calculus 2 Textbook > Differential Equations

Basics of Differential Equations

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Differential Equations

Definitions
A Differential Equations (DE) is an equation that relates a function y(x)y\left(x\right) to some of its derivatives y′, y′′,...y',\ y'',... and other functions in xx.

Examples
y′=y+2sin⁡xy'=y+2\sin x -- first order
y′′′−2y′′− x=2eyy'''-2y''-\ x=2e^y -- third order

  • The Order of a DE is the highest order derivative that appears in that DE
  • If the DE of the form y′=f(y)y'=f\left(y\right) (the independent variable doesn't affect the equation), then it is called autonomous
Some Applicatoins
  • Population growth: dPdt=kP\frac{dP}{dt}=kP
  • Logistic growth: dPdt=kP(1−PM)\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)
  • Motions of spring: d2xdt2=−kmx\frac{d^2x}{dt^2}=-\frac{k}{m}x
  • Newton's law of heating and cooling: dTdt=k(T−Tsurround)\frac{dT}{dt}=k\left(T-T_{\text{surround}}\right)

PAGE BREAK
Solutions to a DE
A General Solution to a DE is a solution y=C...y=C... that involves a constant C (this represents a whole family of solutions)

Example
Show that Ae2x+Be3xAe^{2x}+Be^{3x} is the general solution for the differential equation,   ⁣d2y  ⁣dx2−5  ⁣dy  ⁣dx+6y=0.\frac{\de^2y}{\de{x^2}}-5 \frac{\de{y}}{\de{x}} +6y = 0.
Left hand side of the equation:
d2ydx2−5dydx+6y\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y
=(4Ae2x+9Be3x)−5(2Ae2x+3Be3x)+6(Ae2x+Be3x)=\left(4Ae^{2x}+9Be^{3x}\right)-5\left(2Ae^{2x}+3Be^{3x}\right)+6\left(Ae^{2x}+Be^{3x}\right)
=0=0
This equals the right hand side of the equation.
Therefore, it is a solution for the differential equation.
In fact, it is a general solution because the constants A and B can take on any values (it is not specified).



An Initial Value Problem (IVP) is a DE problem where a side/initial condition is given. In this case, the solution we get is a particular/specific solution

Example
Verify that y=2exy=2e^x is the particular solution to the DE y′=yy'=y with y(0)=2y\left(0\right)=2
Left hand side of the equation:
y′y'
=2ex=2e^x
This equals the right hand side of the equation.
Also, y(0)=2e0=2y\left(0\right)=2e^0=2, so it satisfies the initial condition



An Equilibrium Solution is a solution to a DE such that y′=0y'=0
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Direction Fields/Slope Fields

Sometimes it's helpful to create a graph to show the solutions to a DE -- we sketch short lines representing the value of the derivative at various points (x,y)\left(x,y\right)

Strategies for Sketching Direction Fields

  • Set x=0x=0 (the y-axis): Figure out what happens to the derivatives as y increases (moves up the y-axis) or decreases (moves down the y-axis)
  • Set y=0y=0 (the x-axis): Figure out what happens to the derivatives as x increases (moves right on the x-axis) or decreases (moves left on the x-axis)
  • Find out when the derivative is 0--these correspond to horizontal tangent lines
  • Find other spcecial values of the derivative

Example
Sketch the direction field for the DE y′=x+yy'=x+y, then sketch the particular solution that passes through (0,1)(0,1)



Practice Question

Match the differential equations with the solution graphs below

A.
y′=ex−yy'=e^{x-y}
B.
y′=yx+xsin⁡xy'=\frac{y}{x}+x\sin x
C.
y′=−2xyy'=-2xy
i
ii
iii
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Euler's Method

Instead of solving for a solution to a DE, we can sometimes approximate the solution using Euler's Method.

If you are given the following in the question, you'll be able to approximate the solution at any particular point.
  • dydx=F(x,y)\frac{dy}{dx}=F\left(x,y\right)
  • Initial conidtion y0=y(x0)y_0=y\left(x_0\right)
  • Step size h

How do we approximate the solution to the DE?

1. Create a table

2. Fill out the xnx_n column (xn=xn−1+h\color{orange}x_{n}=x_{n-1}+h)

3. Fill out the yny_n column (yn=yn−1+hF(xn−1, yn−1)\color{orange}y_{n}=y_{n-1}+hF\left(x_{n-1},\ y_{n-1}\right))

Example
Given that y′=2x−yy'=2x-y and y(0)=0y\left(0\right)=0, using a step size of 0.1, estimate y(0.3)y\left(0.3\right).

Therefore, y(0.3)≈0.058y\left(0.3\right)\approx0.058.

Practice: Euler's Method

Use Euler's method with step size 0.5 to approximate the value of y(2)y(2), where y(x)y(x) is the solution of the initial-value problem y′=xy−y2y'=xy-y^2, y(0)=1y(0)=1