cm1902 integration 9 - odes
TRANSCRIPT
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Integration Lecture 9
1st
Order Differential Equations
CM1902 Mathematics 1B
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Ordinary Differential Equations,
ODEs
There are lots of situations in engineering where wecan measure the rate of change of a variable, e.g.
Velocity or acceleration and wish to find an
expression for the underlying independent variable,
e.g. If we know the force on a body we can useNewtons 3rdlaw to find the acceleration but we want to
know the displacement with respect to time.
The above are examples of an Ordinary Differential
Equation There are lots of these equations which are standard
results in engineering maths. They are specified
according to their order. If the equation involves a
first derivative then it is first order, etc.2011-122
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1stOrder ODEs
These are equations involving both a functionsuch as x(t) and its 1stderivative
There are lots of different families of theseequations which depends on the form of f(x, t).
We will look at 2 special cases:
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dxf(x,t)
dt
dxf(t)
dt
dxf(x)g(t)
dt
We can just integrate
each side of this equation
with respect to t
This is trickier!
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Examples 1
Solve the 1storder ODE
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3tdx
6ed t
Solve the 2ndorder ODE 2
2
1 d y1
x dx
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Initial Conditions
To solve a differential equation we must integrateand that introduces an arbitrary constant. That
means that every differential equation has an
infinite number of solutions (an infinite number of
constants).
We usually get a differential equation given along
with a condition that the dependant variable takes a
certain value at a particular value of the
independent variable.
For example, y = 3 when x = 0 or x = 19 when t = 7
Because the condition is often given at x = 0 or t =
0 it is known as an initial condition
The differential e uation is then called an Initial2011-125
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Example 1bSolve the 1storder ODE
subject to the condition x = 3 when t= 0
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3tdx 6ed
t
3tx 2e +CFrom before we had:
This is known asthe GENERAL
solution
This is known as the PARTICULAR solution
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Separation of Variables
Given an equation of the form
We can solve by thinking of as a ratio of
differentials dx and dt. Then
Or more correctly
Now we have two integrals, one in x and one in t.
The variables are separated, mixtures of x and t on
one side or another are not allowed. 2011-127
dx f(x)g(t)dt
dx
dt
1dx g(t)dt
f(x)
1dx g(t)dt
f(x)
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Example 2Solve the 1storder ODE
subject to the condition x = 1 when t= -1
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dxx
dt
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Example 3Solve the 1storder ODE
subject to the condition x(0) = 1
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2dx x cos(t)dt
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Example 4Solve the 1storder ODE
subject to the condition y(2) = 1
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2dy x xydx
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Summary
Differential equations crop up regularly inengineering problems. In years 2 and 3 much of
the mathematics modules are taken up with
obtaining solutions to standard problems
Two methods we have looked at for 1storder
ODEs are
Direct Integration
Separation of Variables
An (initial) condition gives allows us to obtain a
particular solution to an IVP
Now: Tutorial 9 2011-1211