Engineering Mathematics I€¦ · · 2017-03-17Engineering Mathematics I_ 2017 . 2 ... This...
Transcript of Engineering Mathematics I€¦ · · 2017-03-17Engineering Mathematics I_ 2017 . 2 ... This...
4. First-Order Differential Equations (2)
Solutions by Substitutions, Numerical Method
Engineering Mathematics I
Dr. Rami Zakaria
Engineering Mathematics I_ 2017
2
We will discuss 3 special cases where DEs can be turned
into separable equations by variable substitution.
We will accept the following rules via experiment, without
going through proving them.
These rules will help us solving DEs if we couldn’t fit them
into one of the previous categories (separable, linear, etc …)
1. Equations with homogenous functions
3
),(),(),(),(
0),(),(
yxNttytxNandyxMttytxM
dyyxNdxyxM
If we have a DE in the form:
And:
We say that functions M(x,y) and N(x,y) are homogenous functions of the same
degree α.
In other words, we say that a first order ODE is homogeneous id it is possible to write:
Therefore, it is possible to reduce the equation into a separable equation by
substituting:
vyxy
xv
uxyx
yu
or
y
xfy
x
yfy or
4
Example 1, p.62: Solve 0222 dyxyxdxyx
Solution: M(x) and N(x) are homogenous functions of degree 2. define a function u(x) and let
y=ux, then dy=udx+xdu
x
y
cxeyxorx
y
cx
yx
x
y
c
x
x
y
cxx
y
x
y
cxuu
x
dxdu
ux
dxdu
u
u
duuxdxux
uduxduxudxxdxx
xduudxuxxdxxux
22
2
32
3322
22222
ln
ln1ln
lnln1ln2
lnln1ln2
01
210
1
1
0)1()1(
0
0
2. Bernoulli’s Equation
5
1,0;)()( nnyxfyxPdx
dy n
A DE in the form:
is called Bernoulli’s Equation
This equation can be reduced into a linear equation using a substitution u = y1-n .
Example: Solve
Solution:
This is a Bernoulli’s equation which can be reduced using a substitution:
dx
duu
dx
dyuy
yu
3
4
3
1
3
3
1,
x
x
ex
u
dx
due
ueuxdx
duu
3)3(
3
1
3
1
3
4
3
4
3
1
3
4
The equation becomes:
For simplification, multiply both sides by
Now we can solve this equation using an integration factor
Linear equation
cexe
xy
x
cexey
x
cexeudxxexuxexu
dx
d
xeex
xx
xx
xxxx
xdxx
33
3
333
)(
313
1
ln1
),0(; xfor
See also: ex2 p.63
3. Equation of the form:
6
0;)( BCByAxfdx
dy
This equation can be reduced to a separable equation using a substitution: u = Ax + By + C
Example (23.p64): Solve
Solution:
By using a substitution:
21 yxdx
dy
See also: ex3 p.63
7
Previously we have seen that the population growth can be given by the simple relationship
For a more accurate representation, let’s introduce another factor K, which represents the carrying
capacity of the environment. Then we get the relationship:
Then we can write the logistic differential equation of population growth:
Let’s find a solution for this DE.
If you rewrite the equation: We notice that this is a Bernoulli’s equation (n=2)
Pdt
dP
K
PP
dt
dP1
Basically it means, the growth slows
down when P is getting closer to K
bPaPdt
dPor
K
PaP
dt
dP
1 a and b are positive constants.
2bPaPdt
dP
The logistic model of population growth Application
9
Previously we have seen that a simple equation that describes the acceleration of a falling object is
given by:
g
dt
sd
2
2
The velocity of a falling object with air resistance Application
Fr = -bv
W =mg
To improve the model, we add an air resistance force that is proportional to the
velocity Fr = -bv ; b is a positive constant called the drag coefficient.
According to Newton’s second law :
Now, let’s solve this DE:
Rewrite:
dt
dvmbvmgmaF
tm
b
tm
b
ceb
mgv
ecvm
bg
ctvm
bg
b
mdtdv
vm
bg
m
b
b
m
dt
vm
bg
dv
dt
dvv
m
bg
2
1ln
variables theseperate
Exercise: Solve the previous equation for the IC : v(0)=0
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tm
b
tm
b
eb
mgv
b
mgcc
b
mg
vceb
mgv
1
:IVP theofsolution particularA
0
0)0(;
Example for m=20, g=9.8, b=0.75
Euler’s Numerical Method
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So far, we are learning a solution procedure for each different type of equations. The
problem is there will be always different differential equations that we don’t have a
method to solve. Even when we find a solution, we may not be able to write it in an
explicit form.
Also, there are some differential equations that are not easy to graph their direction fields.
In these cases we can use numerical methods that allow us to approximate solutions to
differential equations. There are different methods that can be used, but we are going to
look at one specific method called Euler’s Method.
A first order IVP looks like:
Now we need to approximate the solution of this IVP near t=t0
We know two things here; we know the value of the solution at t=t0 , and we know the
value of the derivative at t=t0 ;
This means that the equation of the tangent line to the solution at t=t0 is:
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If t1 is close enough to t0 then the point
y1 on the tangent line should be very
close to the actual value of the solution
at y(t1)
plug t1 in the equation for the tangent line
Here, we don’t have the value of the solution at y1 and so we won’t know the slope of
the tangent line to the solution at this point. But, as y1 is an approximation to the
solution at t1 , we can use that to estimate the slope of the tangent line at t1 : f(t1, y1).
Therefore the tangent line at t1, can be approximated by:
We take another nearby point t = t2 ; then:
And so on …
Or simply:
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Often, we assume that the step sizes between the points t0 , t1 , t2 , … are of
a uniform size of h. So we can assume that :
Then, the formula for the next approximation becomes:
Example: For the IVP
Use Euler’s Method with a step size of h = 0.1 to
find approximate values of the solution at t = 0.1,
0.2, 0.3, 0.4, and 0.5. Compare them to the exact
values of the solution as these points.
Note: This is a simple linear differential
equation, and you can check that the solution is
Solution:
For Euler’s Method, we rewrite the differential equation into the form:
note that t0 = 0 and y0 = 1
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So, the approximation to the solution at t1 = 0.1 is y1 = 0.9.
At the next step we have
Therefore, the approximation to the solution at t2 = 0.2 is y2 = 0.852967995.
you can check the remainder of these computations:
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tn Approximation yn Actual Value Error
0 1 1 0 %
0.1 0.9 0.925794646 2.79 %
0.2 0.852967995 0.889504459 4.11 %
0.3 0.837441500 0.876191288 4.42 %
0.4 0.839833779 0.876283777 4.16 %
0.5 0.851677371 0.883727921 3.63 %
Note: for more accurate approximation, we can take a smaller step h
See also Ex1. P67