DE Solution Ortho Trajectories Exponential Growth/Decay Differential Equations and Applications Mathematics 54–Elementary Analysis 2 Institute of Mathematics University of the Philippines-Diliman 1 / 22
DE Solution Ortho Trajectories Exponential Growth/Decay
Differential Equations and ApplicationsMathematics 54–Elementary Analysis 2
Institute of MathematicsUniversity of the Philippines-Diliman
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DE Solution Ortho Trajectories Exponential Growth/Decay
Introduction
Definition.
A differential equation is an equation involving a function of onevariable and its derivatives.
Examples.
1dy
dx= 6−3x2
2du
dy= e−yy3
3d2u
dx2 = tanx sec2 x
4 v′′ = 1
x2 +5
Definitions.1 The order of a differential equation is the highest order of the
derivative in the equation. ((1)–(2) are of order 1, (3)–(4) are oforder 2.)
2 A solution of a differential equation is a function y = f (x) thatsatisfies the equation.
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Differential Equations
Consider the differential equation
dy
dx= tan2 3x
y2 −1.
Notice that the above equation can be written as(y2 −1
)dy = tan2 3x dx
Definition
A differential equation that can be expressed in the formf (x)dx = g(y)dy is said to be separable.
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General Solution of a Differential Equation
Example.
1 Find a solution ofdy
dx= cos3x
sin2y.
Solution: We have
sin2y dy = cos3x dx
⇒∫
sin2y dy =∫
cos3x dx
⇒ −1
2cos2y+C1 = 1
3sin3x+C2
.
Combining C1 and C2, the solution has the form
1
2cos2y+ 1
3sin3x = C , C ∈R
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General Solution of a Differential Equation
Example.
2 Solvedu
dx= x
p1−u2
up
2x2 +1.
Solution: The given equation is separable. Indeed,
up1−u2
du = xp2x2 +1
dx
⇒∫
up1−u2
du =∫
xp2x2 +1
dx
⇒−√
1−u2 +C1 = 1
2
√2x2 +1+C2
⇒ C = 1
2
√2x2 +1+
√1−u2.
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Particular Solution of a Differential Equation
Example.
3 Find the equation of the curve that passes through (0,1) andwhose slope at any point
(x,y
)is given by y cosx
1+y2 .
Solution: The equation of the curve satisfiesdy
dx= y cosx
1+y2 .
⇒ 1+y2
ydy = cosx dx
⇒∫ (
1
y+y
)dy =
∫cosx dx
⇒ ln∣∣y∣∣+ 1
2y2 = sinx+C
But (0,1) is a point on the above curve. Thus, we have
ln |1|+ 1
2(1)2 = sin0+C =⇒ C = 1
2.
The equation of the curve is ln |y|+ 1
2y2 = sinx+ 1
2.
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Orthogonal Trajectories
Definition.
An orthogonal trajectory to a set of curves is a curve that intersectseach curve, in the given set, orthogonally (perperdicularly). That is,the tangent lines at the points of intersection are perpendicular.
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Example.
1 Consider the set of circles with center at the origin x2 +y2 = C.
Some orthogonal trajectoriesare:
x−axis,
y−axis,
y = kx, k ∈R
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Computing Equations of Orthogonal Trajectories
Example.
Consider the set of parabolas with vertex at the origin y = kx2.
Note: if each curve in the set hasslope m,then each curve in the set oforthogonal trajectories has
slopemOT =− 1
m.
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Orthogonal Trajectories of y = kx2
Each curve in the set has slope
m = dy
dx= 2kx
Since this is true for all k, andy
x2 = k =⇒ m = 2( y
x2
)x = 2y
xHence, an orthogonal trajectory must have slope mOT =− x
2y.
That is, an orthogonal trajectory must satisfy the differentialequation
dy
dx=− x
2y.
Solving, we obtain y2 =−1
2x2 +C, where C ∈R.
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DE Solution Ortho Trajectories Exponential Growth/Decay
Example.
set of curves : {y = kx2 : k ∈R}set of orthogonal trajectories : {y2 + 1
2 x2 = C : C ∈R}
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DE Solution Ortho Trajectories Exponential Growth/Decay
Orthogonal Trajectories of y = x2 +C
Example.
Consider the set of parabolas y = x2+C, C ∈R, whose vertices are onthe y-axis.
Each curve in the set has slope m = dy
dx= 2x.
Hence, each orthogonal trajectory must have slope
mOT =− 1
m=− 1
2x.
That is, each orthogonal trajectory must satisfy the differentialequation
dy
dx=− 1
2x.
Thus, the solution is y =−1
2ln |x|+K , where K ∈R.
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Example.
set of curves : {y = x2 +C : C ∈R}set of orthogonal trajectories : {y =−1
2 ln |x|+K : K ∈R}
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Orthogonal Trajectories
Example.
Consider the set of curves siny = ex +C.
Computing for the slope m = dy
dxof each curve above :
d
dx
(siny
) = d
dx
(ex +C
)⇒ cosy
dy
dx= ex
⇒ dy
dx= ex secy
Thus, each orthogonal trajectory must satisfy the differentialequation,
dy
dx=− 1
ex secy.
Solving, we obtain ln∣∣secy+ tany
∣∣= e−x +K .14 / 22
DE Solution Ortho Trajectories Exponential Growth/Decay
Mathematical Modelling
Exponential Growth and Decay
A population is said to be growing (or decaying) exponentially if itsinstantaneous rate of increase (or decrease) at any given time isproportional to the population at that time. (e.g., bacterial cultureand radioactive decay)
Let P = P(t) be the population at any time t.
Hence,dP
dtis the rate of change at any time t. Thus,
dP
dt∝ P
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Exponential Growth and Decay
dP
dt∝ P =⇒ dP
dt= kP, k : constant
1
PdP = k dt
lnP = kt +C
P (t) = ekt+C
P (t) = ekteC
Let t = 0 =⇒ P (0) = eC .
Equation of the Model
P = P(t) = P0ekt
where P0 indicates initial population;if k > 0, the model is for exponential growth;
if k < 0, the model is for exponential decay.
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P (t) = P (0)ekt
Example 1.
There are 1000 bacteria at a time and the number doubles after 30minutes. If the population is proportional to its rate of change,(a) find the population of the culture after t hours.(b) How many will be present in 2 hrs?(c) How long will it take before 64000 is present?
(a) The initial population P0 = 1000. Next goal is to find k.
P(1
2
)= 2000 = 1000e(
12 k
)=⇒ e
(12 k
)= 2 =⇒ ek = 4.
Therefore, P (t) = 1000(4)t .
(b) When t = 2, P(2) = 16000.
(c) We find t such that 64000 = 1000(4)t . Hence t = 3 hrs.
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P (t) = P (0)ekt
Example 2.
A bacterial culture triples its population after an hour. If 81M werepresent at the end of 4hrs, how many were present initially.
We must find P0 = P (0). Note that P (1) = 3P0 = P0ek·1 ⇒ ek = 3.Hence we have
P(t) = P0(3)t
81M = P (4) = P034
Hence, P0 = 1M .
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P (t) = P (0)ekt
"Radioactive substances decay by spontaneous emission of radiation. It’s
been experimentally found that the rate of decay is proportional to the
amount of the substance at any given time".
The half-life of a substance is the time it takes for half of theamount to decay.
Example 3.
Element X has a half life of 3 yrs. If 64g of it is present at a time, howmuch of it will be present after 16 yrs.
Half-life of 3yrs ⇒ P (3) = 12 P0 = P0e3k ⇒ e3k = 1
2 ⇐= ek = (12
)1/3.
Find P (16):
P (16) = P0e16k = 64(ek
)16 = 64(1
2
) 163 = 64
3p
216= 64
32 3p
2= 3p
4 g.
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P (t) = P (0)ekt
Example 4.
30% of a radioactive substance disappears in 15yrs. Find its half-life.
Find t such that P (t) = 12 P0: P (15) = (0.70)P0.
P (15) = P0e15k ⇒ P0e15k = 0.7P0
⇒ e15k = 0.7
⇒ ek = (0.7)1
15 .
We now want to know at t is
P (t) = 12 P0 =⇒ 1
2 P0 = P0ekt
=⇒ 12 = (
ek)t = (0.7)
t15
=⇒ t = −15ln2
ln0.7.
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Exercises
1 Solve the following differential equations.
1dy
dx= xex
y√
1+y2
2dy
dx= ey sin2 x
y secx
3dy
dx= y cosx
1+y2 , y (0) = 1
4 xy′+y = y2, y (1) =−1
2 Do as indicated.1 Find the equation of the curve that passes through the point
(0,1) and whose slope at(x,y
)is xy.
2 Find the function f such that f ′ (x) = f (x)(1− f (x)
)and f (0) = 1
2.
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Exercises
3 Find the orthogonal trajectories of the family of curves.1 x2 +2y2 = K2 y2 = Kx3
3 y = x
1+Kx4 Do as indicated.
1 A bacteria culture starts with 500 bacteria. After 3 hrs there are8000 bacteria. Find: (i) the initial population; (ii) the formulafor the population after t hours; (iii) the number of bacteriaafter 5 hours; (iv) the rate of growth after 5 hours; (v) the timewhen population reaches 200000.
2 Bismuth-210 has a half-life of 5 days. At start, there is an 800mgsample. Find: (i) a formula for the mass remaining after t days;(ii) mass remaining after 30 days; (iii) time when the remainingamount is 1mg.
3 After 3 days a sample of Radon-222 decays to 58% of its originalamount. Find: (i) the half-life of Radon-222; (ii) the time itwould take for the sample to decay 10% of its amount.
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