application of first order ordinary Differential equations

EMDADUL HAQUE MILON
ehmilon24171@gmail.com
mdehmilon24171@gmail.com
DEPERTMENT OF STATISTICS,
UNIVERSITY OF RAJSHAHI..
RAJSHAHI, BANGLADESH
6205&6206

DEPARTMENT OF
statistics, UNIVERSITY OF
RAJSHAHI.
2Group D

Welcome to all of our
presentation of
GROUP-D
3Group D

1. Md. Jenarul Islam
1510224120
2. Md. Lukman Joni
1510924175
3. Md. Shamim Hossen
15010424107
4Group D

4. Md. Sharifuzzaman
1510224164
5. Md. Amdadul Hasan
1510924174
6. Md. Emdadul Haque (Milon)
1510924171
5Group D

7. Md. Abdul Bari
1510324161
8. Md. Naoaj Sharif
1510524171
9. Md. Shahinur alam
1510824130
6Group D

10. Md. Abdul Alim
1510824155
7Group D

TOPICS OF OUR
PRESENTATION
8Group D

1. Applications of first order
ordinary differential
equation.
2. Orthogonal trajectory. &
3. Oblique trajectory.
9Group D

Applications of 1st order
ordinary differential
equation.
1.
10Group D

1st order ordinary differential equation:
Definition of 1st order ordinary differential equation: 1st order ordinary
differential equation is one kind of differential equation. A differential
equation involving ordinary derivatives of one or more dependent variables
with respect to a single independent variable and which has only one order
derivatives, is called a 1st order ordinary differential equation.
Example:
1.
𝑑𝑦
𝑑𝑥
=
𝑥2+𝑦2
𝑥+𝑦
is a 1st order ordinary differential equation. Here y
is a dependent variable and x is a independent variable and 𝑑
𝑑𝑥 is the
derivative term which order is one, so it is a 1st order ordinary differential
equation.
11Group D

Standard form of 1st order ordinary
differential equation:
The standard form of 1st order ordinary differential equation is
𝑑𝑦
𝑑𝑥
= 𝑓 𝑥, 𝑦 ⋯ ⋯ ⋯ ⋯ ⋯ 1
or the differential form
𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0 ⋯ ⋯ ⋯ 2
In the form (1) it is clear from the notation itself that y is regarded as
the dependent variable and x as the independent one.
But in the form (2) we may actually regard either variable as the
dependent one and the other as the independent.
12Group D

Applications of 1st order ordinary differential equation :
There are a lot of applications of 1st order
ordinary differential equation in our real life
in various sectors. Some of these are given
below:
13Group D

1. Cooling/Warming Law (use in physics)
2. Population Growth and Decay (in stat..)
3. Radio-Active Decay and Carbon Dating
4. Mixture of Two Salt Solutions(in
chemistry)
5. Series Circuits (in physics)
6. Survivability with AIDS (in medicine)
7. Draining a tank (in engineering)
14Group D

8. Economics and Finance ( in economics)
9. Mathematics Police Women
10. Drug Distribution in Human Body ( in
biology)
11. A Pursuit Problem
12. Harvesting of Renewable Natural
Resources
(in agriculture)
13.Determining the motion of a projectile, rocket,
satellite or planet (in engineering).
14.Determining the charge or current in a
electric
15Group D

15. Determination of curves that have certain
geometrical properties.
16. Conduction of heat in a rod or in a slab.
17. Determining the vibrations of a wire or
membrane.
And so on.
16Group D

Some applications of 1st order ordinary
differential equation in engineering
17Group D

Falling
stone
Parachute
Water
level tank
Vibrating
spring
Beats of vibrating system
Current
circuit
Pendulum
Prey model
18Group D

Lets see some applications
of 1st order ordinary
differential equation with
example.
19Group D

1. Population Growth and
Decay (in statistics)
Problem: A population grows at the rate of 5% per year. How long does it takes for the
population to double?
Solution: Let the initial population be p0 and let the population after t years
be p.
Then we get,
𝑑𝑝
𝑑𝑡
=
5
100
𝑝
⇒
𝑑𝑝
𝑑𝑡
=
𝑝
20
⇒
𝑑𝑝
𝑝
=
1
20
𝑑𝑡 ⋯ ⋯ ⋯ ⋯ ⋯ (1)
which is a separable differential equation.
Now we integrating (1) and we get,
20Group D

⇒ 𝑙𝑛𝑝 =
𝑡
20
+ 𝑐 ⋯ ⋯ ⋯ ⋯ (2)
We put the initial value in (2) i.e. 𝑡 = 0 and 𝑝 = 𝑝0
𝑙𝑛𝑝0 = 𝑐
⇒𝑐 = 𝑙𝑛𝑝0 ⋯ ⋯ ⋯ ⋯ ⋯ (3)
We get from (2) and (3),
⇒ 𝑙𝑛𝑝 =
𝑡
20
+ 𝑙𝑛𝑝0
⇒𝑡 = 20ln(
𝑝
𝑝0
)
21Group D

when, p=2p0 then,
⇒ 𝑡 = 20𝑙𝑛(
2𝑝0
𝑝0
)
= 20𝑙𝑛2
= 13.86 ≈ 14 years
Hence the population is double in 14 years.
22Group D

2. Determination of curves that have certain
geometrical properties.
Problem: The slope of the tangent at a point (x, y) on a curve is − 𝑥
𝑦 .
If the curve passes through the point (3,-4). Find the equation of the curve.
Solution: We know the slope of a curve at point (x, y) is
𝑑𝑦
𝑑𝑥
Thus by the problem
𝑑𝑦
𝑑𝑥
=
−𝑥
𝑦
⇒ 𝑦𝑑𝑦 + 𝑥𝑑𝑥 = 0 ⋯ ⋯ ⋯ ⋯ ⋯ (1)
Which is a separable differential equation.
23Group D

Now integrating (1) and we get,
𝑦2
2
+
𝑥2
2
= 𝑐
⇒𝑥2 + 𝑦2 = 2𝑐 ⋯ ⋯ ⋯ ⋯ (2)
where c is an arbitrary constant.
Since the curve passes through the point (3, -4)
thus from (2),
32 + (−4)2= 2𝑐
⇒𝑐 =
25
2
24Group D

Hence,
𝑥2
+ 𝑦2
= 25,
which is the required equation of the curve.
25Group D

3. Cooling/Warming Law (use in physics)
Example : When a chicken is removed from an oven, its temperature is
measured at 3000F. Three minutes later its temperature is 200o F. How
long will it take for the chicken to cool off to a room temperature of
70oF.
Solution: From the given problem ,
Tm = 70 and T=300 at for t=0.
T(0)=300=70+c2e.0
This gives c2=230
For t=3, T(3)=200
Now we put t=3, T(3)=200 and c2=230 then
26Group D

230
1303αe 
23
13
ln3 
19018.0
23
13
ln
3
1

27Group D

Thus T(t)=70+230 e-0.19018t
We observe that furnishes no finite solution to T(t)=70
since
limit T(t) =70.
t 
We observe that the limiting temperature is 700F.
28Group D

4. Carbon Dating
Example. Find the age of an object that has been excavated and found
to have 90% of its original amount of radioactive Carbon 14.
Solution: Using the equation y = y0ekt we see that we must ﬁnd two
things:
(i) the value of k
(ii) the value of t
for which y0ekt = (90/100)y0, i.e., ﬁnd t such that ekt = (9/10).
(i) Find the value of k:
Rearranging the half-life equation and using the fact that the half-life is
known to be 5570 years, we have
29Group D

⇒−k =(ln2/half-life)
⇒-k =ln2/5570 ≈ .0001244
So k = −.0001244.
(ii) Find the value of t which
makes ekt = 9/10:
⇒ e−.0001244t = 0.9
⇒−.0001244t = ln(0.9)
⇒ t = [ln(0.9)/−.0001244] ≈ 878
Conclusion :
Therefore the sample is approximately 878 years old.
30Group D

5.Radioactive Decay:
Radioactive decay: radioactive decay is the process of losing nuclei….
Half-life: The Half-life of a radioactive element is the time required for half of
the radioactive nuclei decay present in a sample (i.e. for the quantity to be
reduced by one-half).
The half-life of an element is totally independent of the number of nuclei
present initially, because this decay occurs exponentially, i.e. according to the
diﬀerential equation dy/dt = ky, where k is some negative constant. (The value
of the constant diﬀers for the various radioactive elements.)
• Let y0 be the number of radioactive nuclei present initially. Then the number
y of nuclei present at time t will be given by:
31Group D

⇒y = y0ekt
Since we are looking for the half-life, we wish to know
the time t at which only
1/2(y0) nuclei remain:
⇒Y0ekt =1/2(y0)
⇒ ekt =1/2
32Group D

⇒ kt = ln(1/2) = ln1−ln2 = −ln2
⇒ t =ln2 −k
(Since k is negative, ln2 −k is positive.) Thus the half-life depends only
on k.
The formula above is worth noting for future use:
half-life = ln2 −k
33Group D

Example 13. The number of atoms of plutonium-210 remaining after t
days, with an initial amount of y0 radioactive atoms, is given by:
y = y0e(−4.95×10−3)t
Find the half-life of plutonium-210.
Solution: We see that for this element we have k = −4.95×10−3. Using the
formula above we have:
Half-life =(ln2/−k)={(ln2)/(4.95×10^−3)} ≈ 140
Therefore the half-life of plutonium-210 is approximately 140 days.
34Group D

2. Orthogonal
Trajectory
35Group D

Orthogonal Trajectory:
Definition: An orthogonal trajectory of a family of curves is a
curve that intersects each curve of the family orthogonally, that is, at
right angles (see Figure 7).
Figure:7 36Group D

Another Definition: Let
𝐹 𝑥, 𝑦, 𝑐 = 0 ⋯ ⋯ ⋯ ⋯ 2.1
be a given one-parameter family of curves in the 𝑥𝑦 plane. A curve
that intersects the curves of the family (2.1) at right angles is called an
orthogonal trajectory of the given family.
Example: Consider the family of circles
𝑥2 + 𝑦2 = 𝑐2 ⋯ ⋯ ⋯ ⋯ ⋯ 2.2
with center at the origin and radius c. Each straight line through the
origin,
𝑦 = 𝑘𝑥, ⋯ ⋯ ⋯ (2.3)
37Group D

is an orthogonal trajectory of the family of circles (2.2).
Conversely, each circle of the family (2.2) is an orthogonal trajectory
of the family of the straight lines (2.3). So the families (2.2) and (2.3)
are orthogonal trajectories of each other.
In figure (2.a.1) several members of the family of circles (2.2), drawn
solidly and several members of the family of straight lines (2.3),
drawn with dashes are shown.
38Group D

X
Y
Figure (2.a.1)
Member of the family of
circles (2.2)
Member of the family
of straight lines (2.3)
39Group D

Procedure for finding the orthogonal
trajectories of a given family of curves:
Step 1. From the equation
𝐹 𝑥, 𝑦, 𝑐 = 0 ⋯ ⋯ ⋯ ⋯ 2.4
of the given family of curves, at first we find the differential
equation
𝑑𝑦
𝑑𝑥
= 𝑓 𝑥, 𝑦 ⋯ ⋯ ⋯ ⋯ ⋯ 2.5
of this family.
Step 2. In the differential equation (2.5), so found in Step 1, replace f(x, y) by its
negative reciprocal −1 𝑓(𝑥, 𝑦) . This gives the differential equation
40Group D

𝑑𝑦
𝑑𝑥
=
−1
𝑓(𝑥,𝑦)
⋯ ⋯ ⋯ ⋯ ⋯ 2.6
of the orthogonal trajectories.
Step 3. At last we obtain a one parameter family
G x, y, c = 0 or 𝑦 = 𝐹(𝑥, 𝑐)
of solution of the differential equation (2.6), thus obtaining the
desired family of orthogonal trajectories (except possibly for
certain trajectories that are vertical lines and must be
determined separately).
Caution: In step 1, in finding the differential equation (2.5) of the given
family, be sure to eliminate the parameter c during the process.
41Group D

Example of procedure of finding Orthogonal
Trajectory:
Example: Find the orthogonal trajectories of the family of parabolas
𝒚 = 𝒄𝒙 𝟐
Solution: Given the family of parabolas
𝑦 = 𝑐𝑥2 ⋯ ⋯ ⋯ ⋯ (2.7)
Step 1. We first find the differential equation of the given family (2.7)
Differentiating (2.7), we obtain
𝑑𝑦
𝑑𝑥
= 2𝑐𝑥 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2.8)
Eliminating the parameter c between equation (2.7) and (2.8)
42Group D

∴
𝑑𝑦
𝑑𝑥
=
2𝑦
𝑥
⋯ ⋯ ⋯ ⋯ ⋯ (2.9)
which is the differential equation of the family (2.7).
Step 2. we now find the differential equation of the orthogonal
trajectories by replacing
2𝑦
𝑥
in (2.9) by its negative reciprocal, obtaining
𝑑𝑦
𝑑𝑥
= −
𝑥
2𝑦
⋯ ⋯ ⋯ ⋯ ⋯ (2.10)
Step 3. we now solve the differential equation (2.10).
Here,
2𝑦𝑑𝑦 = −𝑥𝑑𝑥
Which is separable differential equation. Now integrating and we get,
𝑥2 + 2𝑦2 = 𝑘2
43Group D

Where k is an arbitrary constant.
This is the family of orthogonal trajectories of (2.7), it is clearly a
family of ellipses with centers at the origin and major axes along
the X axis.
Some members of the original family of parabolas and some of the
orthogonal trajectories (ellipses) are shown in figure (2.a.2).
44Group D

X
Y
Member of the
orthogonal
trajectories (ellipses) Member of the original
family of parabolas
Figure (2.a.2) 45Group D

3. Oblique Trajectory
46Group D

Oblique trajectory:
Definition: Let
𝐹(𝑥, 𝑦, 𝑐) = 0 ⋯ ⋯ ⋯ ⋯ (3.1)
be a one parameter family of curves. A curve that intersects
the curves of the family (3.1) at a constant angle ∝≠ 90° is
called an oblique trajectory of the given family.
47Group D

Procedure for finding the oblique trajectories of a given family of
curves:
Let
𝐹 𝑥, 𝑦, 𝑐 = 0 ⋯ ⋯ ⋯ ⋯ (3.2)
be a family of curves.
Suppose the slope of (3.2) is
𝑑𝑦
𝑑𝑥
= 𝑓 𝑥, 𝑦 ⋯ ⋯ ⋯ ⋯ ⋯ 3.3
Let Ѳ be the angle of X axis and the tangent of the given family of
curves. Thus,
tan 𝜃 = 𝑓(𝑥, 𝑦)
48Group D

⇒𝜃 = tan−1 𝑓(𝑥, 𝑦)
Again let the angle of the tangent of oblique trajectories and X axis be
𝜑.
Now, by the theorem of triangle, we get
𝜑 = 𝜃 + 𝛼 ; where 𝛼 is the angle between two
triangles.
⇒𝜑 = tan−1 𝑓 𝑥, 𝑦 + 𝛼
⇒tan 𝜑 = tan[tan−1 𝑓 𝑥, 𝑦 + 𝛼]
=
tan tan−1{𝑓(𝑥,𝑦)}+tan 𝛼
1−tan tan−1 𝑓 𝑥,𝑦 .tan 𝛼
=
𝑓 𝑥,𝑦 +tan 𝛼
1−𝑓 𝑥,𝑦 .tan 𝛼
50Group D

∴ tan 𝜑 =
Therefore, the slope of oblique trajectories is
𝑑𝑦
𝑑𝑥
=
which is a differential equation.
And after solving the equation, we have to get the function of oblique
trajectories of the given family of curves.
51Group D

Example of procedure of finding oblique
trajectory:
Example: Find a family of oblique trajectories that intersects the family of
straight lines
𝑦 = 𝑐𝑥
at angle 45°.
Solution: Given the family of straight lines
𝑦 = 𝑐𝑥 ⋯ ⋯ ⋯ ⋯ (3.4)
Differentiating, we obtain,
𝑑𝑦
𝑑𝑥
= 𝑐 ⋯ ⋯ ⋯ ⋯ (3.5)
Putting the value of c from (3.4) in (3.5) and we get,
𝑑𝑦
𝑑𝑥
=
𝑦
𝑥
⋯ ⋯ ⋯ ⋯ ⋯ (3.6)
52Group D

which is the differential equation of the given family of straight lines.
Now, we replace 𝑓 𝑥, 𝑦 =
𝑦
𝑥
in equation (3.6) by
=
𝑦
𝑥+tan 45°
1− 𝑦
𝑥∙tan 45°
; putting, 𝛼 = 45°
=
𝑦
𝑥+1
1− 𝑦
𝑥
=
𝑥+𝑦
𝑥−𝑦
Thus the differential equation of the desired oblique trajectories is
𝑑𝑦
𝑑𝑥
=
𝑥+𝑦
𝑥−𝑦
⋯ ⋯ ⋯ ⋯ ⋯ (3.7)
Now, we solve the differential equation (3.7) and we get,
53Group D

𝑥 − 𝑦 𝑑𝑦 = 𝑥 + 𝑦 𝑑𝑥
⇒ 𝑥 − 𝑦 𝑑𝑦 − 𝑥 + 𝑦 𝑑𝑥 = 0
which is the homogeneous differential equation. Now we write this in
𝑔( 𝑦
𝑥) form we get,
𝑑𝑦
𝑑𝑥
=
𝑥+𝑦
𝑥−𝑦
=
1+ 𝑦
𝑥
1− 𝑦
𝑥
⋯ ⋯ ⋯ ⋯ ⋯ 3.8
Let, 𝑦 = 𝑣𝑥
∴
𝑑𝑦
𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (3.9)
Now from (3.8) and (3.9) we get,
54Group D

𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
=
1+𝑣
1−𝑣
⇒ 𝑥
𝑑𝑣
𝑑𝑥
=
1+𝑣
1−𝑣
− 𝑣
⇒ 𝑥
𝑑𝑣
𝑑𝑥
=
𝑣2+1
1−𝑣
⇒ 𝑥𝑑𝑣 =
𝑣2+1
1−𝑣
𝑑𝑥
⇒
1−𝑣
1+𝑣2 𝑑𝑣 =
1
𝑥
𝑑𝑥
⇒ −
𝑣
𝑣2 + 1
𝑑𝑣 +
1
𝑣2 + 1
𝑑𝑣 =
1
𝑥
𝑑𝑥
⇒−
1
2
𝑙𝑛 𝑣2
+ 1 + tan−1
𝑣 = 𝑙𝑛 𝑥 + 𝑙𝑛(𝑐)
⇒ ln 𝑣2 + 1 − 2 tan−1 𝑣 = −𝑙𝑛 (𝑐𝑥)2
55Group D

⇒ln
𝑦2
𝑥2 + 1 + 𝑙𝑛𝑥2 𝑐2 − 2 tan−1 𝑦
𝑥 = 0
⇒ln
𝑦2+𝑥2
𝑥2 + 𝑙𝑛𝑥2
𝑐2
− 2 tan−1 𝑦
𝑥 = 0
⇒ln 𝑐2 𝑦2 + 𝑥2 − 2 tan−1 𝑦
𝑥 = 0
Which is the family of oblique trajectories.
56Group D

1.Class lecture - Prof. Dr. Md. Golam Hossain, department of Statistics
University of Rajshahi.
2. Ross, S.L.(1989). Differential Equations, 4th ed., Wiley, N.Y.
3. Google search:
https://www.google.com.bd/search?q=ENGG2013+Unit+24&oq=ENGG2013+Unit+24&aqs=chrome..69i
57.2731j0j7
&sourceid=chrome&ie=UTF-8
57.
2731j0j7&sourceid=chrome&ie=UTF-8#q=minggu-2-1-engineering-mathematics-differential-equations
57.
2731j0j7&sourceid=chrome&ie=UTF-8#q=StewCal4e_7_3+ppt
57Group D

application of first order ordinary Differential equations

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