DIFFERENTIAL EQUATIONS
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats)
April 7, 2017
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
ORDINARY DIFFERENTIAL EQUATIONS
In many physical situation, equation arise which involve differentialcoefficients. For example:
1 The rate of decrease of temperature of a hot body.
2 A body falling freely under gravity.
3 The oscillation on the end of a spring.
4 The decay of a radioactive substance.
5 The decrease in concentration of chemical compound in a reaction.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
FORMATION OF DIFFERENTIAL EQUATIONS
Differential equations arise or may be derived in a variety of ways. Inmost cases the problem is to find the dependent variable in terms of theindependent one. Eg. Finding x in terms of t or y in terms of x .Differential equations may be formed by direct differentiation.
Eg. y = x3 + 7x2 + 3x + 7 −→Eqn 1
dydx = 3x2 + 14x + 3 −→Eqn 2
d2ydx2 = 6x + 14 −→Eqn 3
d3ydx3 = 6 −→ Eqn 4
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Definition
Equations which contain an independent variable, a dependent variableand at least one of their derivatives are called DIFFERENTIALEQUATIONS. Thus a relationship between a variable quantity x and a
dependent function y and its derivatives dydx ,
d2ydx2 ,
d3ydx3 · · · is called an
ORDINARY DIFFERENTIAL EQUATION.Equations 2,3 and 4 are differential equations.
Other examples of ordinary differential equations
1dydx = kx
2 x2(1 + y) dydx − (1 + x)y2 = 0
3d2ydx2 = −n2y
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
4. ( dydx )2 + 3y = 0
5. x d3ydx3 + d2y
dx2 + x( dydx )4 = 0
KINDS OF DIFFERENTIAL EQUATIONS
There are two main types of differential equatons.
Ordinary Differential Equations(O.D.E.’s) These are differentialequations with only one independent variables.
Partial Differential Equations(P.D.E’s) These are differentialequations with more than one independent variable.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Examples of partial differential equations
x ∂z∂y + y ∂z∂x = 0
∂2f∂y2 = 2x
x2+y2
∂2z∂x2 + ∂2z
∂y2 = 0
fxx + fyy = 0
∂2z∂x2 + ∂2z
∂y2 = ∂z∂t
NOTE:
This course is concerned more with ordinary differential equations.Partialdifferential equations may be considered slightly.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
ORDER OF A DIFFERENTIAL EQUATION
Differential equations are classified according to the highest derivativewhich occurs in the equations.Thus if the highest derivative that occursin an equation dny
dxn ,the equation is said to be of order n.
On slide 3, we notice that:
Equation 2 is of the first order ,having only the first derivative.
Equation 3 is of the second order.
Equation 4 is of the third order.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
DEGREE OF A DIFFERENTIAL EQUATION
The degree of a differential equation is the highest power of the highestderivative which the equation contains.
Thus( d2ydx2 )3 + 3 dy
dx = 0 is of the second order and third degree.Considering the examples on slide 4 and 5, we also notice that example1,2,3 and 5 are of first degree whilst example 4 is of a second degree.NOTE that in example 5 the degree of the equation is determined by the
power of the highest derivative d3ydx3 and not by the fourth power term in
dydx
Exercises
Indicate the order and the degrees of the following ordinary differentialequations.
1. ( d2ydx2 )3 + dy
dx = y − x 2. x2 dydx + x d4y
dx4 = 3y
3.( dydx )4 + xy = x 4. ex dy
dx = (1− yex)
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
SOLUTIONS OF A DIFFERENTIAL EQUATION
We recall that an ordinary differential equation was defined as arelationship between a variable quantity x and a dependent function yand its derivatives.These equations normally arise from physical situations and it is oftenrequired to obtain a functional relationship between x and y alone,having eliminated the derivatives. This relation is referred to theSOLUTION of the differential equation.
A solution which is COMPLETE or GENERAL must contain a number ofarbitrary constants which is equal to the order of the equation. Solutionsof the differential equation with the appropriate number of arbitraryconstants are called GENERAL SOLUTIONS.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
In physical problems, solutions are usually required which satisfy certainspecified conditions. These provide information from values to beassigned to the arbitrary constants.
This type of solution , which satisfies certain definite conditions, is calleda PARTICULAR SOLUTION and the conditions satisfied are calledBOUNDARY CONDITIONS or INITIAL CONDITIONS.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example
Considerdy
dx= x which is of the first order.
Integrating we have∫dy =
∫xdx
y =1
2x2 + A
Now if we consider the general solutions y =1
2x2 + A to the equation
dy
dx= x .
Let us assume that the boundary condition is given to be y = 1 whenx = 0⇒ A = 1
⇒ The value assigned to A = 1 and the particular solution is y =1
2x2 + 1
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
DIFFERENTIAL EQUATIONS OF THE FIRST ORDERAND FIRST DEGREE
Let us first look at the case where one variable is absent.
(a) When y is absent
The general form isdy
dx= f (x) ⇒
∫dy =
∫f (x)dx
y =∫f (x)dx
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example
Solve the differential equationdy
dx= x4 + sin x
⇒∫dy =
∫(x4 + sin x)dx
y =1
5x5 − cos x + c
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
(b) When x is absent
The general form isdy
dx= f (y)⇒ dy = f (y)dx Rewriting this in the form :
dx
dy=
1
f (y)∫dx =
∫ 1
f (y)dy =
∫ dy
f (y)
Example
Solve the equationdy
dx= tan y
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solution
dy
dx= tan y
dx
dy=
1
tan y=
cos y
sin y
x =
∫cos y
sin ydy
= ln | sin y | +c
The examples given above lead us to the main types of first order, firstdegree differential equations and their solutions.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
TYPES OF FIRST ORDER DIFFERENTIAL EQUATIONS
1 Variables Separable
2 Homogeneous
3 Linear
4 Exact
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
TYPE 1 - VARIABLES SEPARABLE
If the terms of the equation can rearranged into two groups, eachcontaining only one variable, the variables are said to be SEPARABLE.Since the differential equations of the first order and first degree containdy
dxto the first power only, they can be written as
dy
dx= F (x , y)
In many cases F (x , y) may be written as
F (x , y) = f (x)g(y)
where f (x) and g(y) are functions of x only and g(y) is a function of yonly.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
We may then ”separate the variables ” and write.
dy
g(y)= f (x)dx
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Worked Example On Type 1
Example
Solvedy
dx=
5x
7y
Solution
The variables are separable
7ydy = 5xdx
⇒∫
7ydy =
∫5xdx
7
∫ydy = 5
∫xdx
7 · 1
2y2 = 5 · 1
2x2 + C ⇒ y2 =
5
7x2 + C
Note : C = C1 + C2 and C is an arbitrary constant.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Exercise 1
Solve xdx + ydy = xy(xdy − ydx)
Solution
x + ydy
dx= xy
(xdy
dx− y
)dy
dx(y − x2y) = −(x + xy2) = −x(1 + y2)∫
ydy
1 + y2= −
∫xdx
1− x2
1
2ln |1 + y2| =
1
2ln |1− x2|+ lnC1
(1 + y2)12 = C1(1 + x2)
12
(1 + y2) = C (1 + x2)Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Exercise 2
Solvedy
dx= (x + y)2
Solution
Let z = x + y
dz
dx= 1 +
dy
dx(1)
=⇒ dy
dx= z2 (2)
From (1)dy
dx=
dz
dx− 1
=⇒ dz
dx− 1 = z2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solution Cont’d
dz
dx= z2 + 1∫
dz
z2 + 1=
∫dx
tan−1 z = x + c
z = tan(x + c)
x + y = tan(x + c)
y = tan(x + c)− x
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solve the following equations
1 idy
dx= tan2(x + y)
iidy
dx= (x + 4y)2
2 i (1 + x)y + (1− x)ydy
dx= 0
iidy
dx+
k
x2= 0
3 i (x + 1)dy − ydx = 0ii (y 2 − x2) + 2xydx
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
HOMOGENEOUS
TYPE 2 - - HOMOGENEOUS
M(x , y) is said to be a homogeneous function of degree n if the sum ofthe powers of x and y in each term of M is n
Eg. (i) x2y − 3xy2 + 2y3 is homogeneous of degree 3(ii) x4 − 2x2y2 is homogeneous of degree 4.
If a first order D.E. is written in the form
dy
dx=
M(x , y)
N(x , y),
where M and N are homogeneous functions of the same degree, then theequation is said to be HOMOGENEOUS.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Examples:
idy
dx=
xy
x2 + y2
ii (x2 + y2)dy
dx= xy
Check
i. (x2 + y)dy
dx= xy ?
ii.dy
dx=
y(3x2 + y2)
x(x + 3y)?
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
METHOD OF FINDING SOLUTION TO AHOMOGENEOUS DIFFERENTIAL EQUATION
If, in the equation;dy
dx=
M(x , y)
N(x , y), (3)
both M and N are homogeneous of degree n, we may divide them by xn
and express the R.H.S as a function of the single variable v , where v =y
x
⇒ y = vx
dy
dx= v + x
dv
dx(4)
Substituting (4) in the differential equation (3), we find that the result isa new differential equation in which the variables v and x and finally
replace v byy
x.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example 1
(x2 + y2)dy
dx= xy
Let v =y
x
y = vx
dy
dx= v + x
dy
dx−−−−− (1)
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Now (x2 + y2)dy
dx= xy
dy
dx=
xy
x2 + y2=
yx
1 + ( yx )2
dy
dx=
v
1 + v2−−−−− (2)
From (1) and (2)
v + xdv
dx=
v
1 + v2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
xdv
dx=
v
1 + v2− v =
−v3
1 + v2
xdv
dx=−v3
1 + v2
Now we notice that x and v are separable. So we separate the variables,integrate and substitute for v to obtain the general solution.
Thus xdv
dx=−v3
1 + v2∫dx
x=
∫(− 1
v3− 1
v)dv
ln(x) =1
2v2− ln(v) + ln(A)
where ln(A) = C
logey
A=
x2
2y2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
y
A= e
x2
2y2
y = Aex2
2y2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example 2
Example 2
dy
dx=
y(3x2 + y2)
x(x2 + 3y2)
Let v =y
xy = vx
dy
dx= v + x
dy
dx
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
v + xdy
dx=
yx (3 + ( y
x )2)
1 + 3( yx )2
v + xdy
dx=
v(3 + v2)
1 + 3v2∫dx
x=
∫1 + 3v2
2v(1− v2)dv
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
ln(x) =1
2
∫ [1
v+
2
1− v− 2
1 + v
]dv
Finally, we have
loge x2 = loge
Av
(1− v2)2
⇒ (x2 − y2)2 = Axy
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Linear Type of Ordinary Differential Equations
If a differential equation can be written in the formdy
dx+ Py = Q, where
P and Q are functions of x only, the equation is said to be LINEAR of
the first order sincedy
dxand y occurs linearly.
Examples:
1dy
dx+ 2y cot x = cos x
2dy
dx+
x
1 + x2y =
1
2x(1 + x2)
3dy
dx− x
(1− x2y =
1
(1− x2
4dy
dx+ y tan x = sec x
5dy
dx− 2xy = 2x
6dy
dx+ xy = x
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
METHODS OF SOLUTION
In the standard linear equation
dy
dx+ P(y) = Q,
the presence of the termsdy
dxand y suggests the differentiation of a
product involving y .To produce this product we multiply the equation throughout by afunction u to be determined later. Thus we have
udy
dx+ uP(y) = uQ
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
The equation reduces todu
dx= uP∫
du
u=
∫Pdx
loge u =
∫Pdx
u = e∫Pdx
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
We note that no arbitrary constants needs to be included here since theconstant required in the solution of the original differential equation willarise on performing the integration.The function
u = e∫Pdx
is referred to as the INTEGRATING FACTOR (I.F)
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example 1
Solve xdy
dx+ 2y = ex
Solution
xdy
dx+ 2y = ex
Writing in standard form yields
dy
dx+
2
xy =
ex
x−−−−(∗)
⇒ P =2
x
⇒∫
Pdx =
∫2
xdx = 2 ln(x) = ln(x)2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solution Cont’d
Thus
u = e∫Pdx = e
∫2x dx
e loge(x)2
= (x)2
Multiplying (*) by u = x2, we have
x2dy
dx+ 2xy = xex
d(x2y)
dx= xex
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
x2y =
∫xexdx
x2y = (x − 1)ex + c1
y =ex(x − 1)
x2+
c1x2
y =ex(x − 1)
x2+ c2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Example 2
Solve the equation
cos xdy
dx+ y sin x = 1
Solution
Writing in standard form:
dy
dx+
sin x
cos xy = sec x −−−−− (∗)
I.F, u = e∫
sin xcos x dx = e
∫tan xdx
Now
∫sin x
cos xdx = ln | cos x | = ln | sec x |
I.F, u = e ln | sec x| = ln | sec x |
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solution
Multiplying through (*) by the I.F = ln | sec x |,we have
sec xdy
dx+
sin x
cos x(sec x)y = sec x sec x
sec xdy
dx+ tan x(sec x)y = sec2 x
⇒ y · sec x =
∫sec2 x = tan x + c1
⇒ y =tan x
sec x+
C1
sec x⇒ y = sin x + c2
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Exercises
Solve the following
1 (x2 + 1)dy
dx+ 2xy = 4x2 given that when
x=3, y=4
(I.F= (x2 + 1) Soln: y = 4x3
x2+1 + 4x2+1
)2 x(1− x2) dy
dx + (2x2 − 1)y = x3
(I.F = 1x√1−x2
; y=x + A1x√
1− x2 = x + A2)
3dydx = y − x(y = x + 1 + cex)
4 tan x dydx = 1 + y
(y + 1 = c sin x)
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
SOME APPLICATIONS OF DIFFERENTIAL EQUATIONS
We recall that a derivative is a rate of change. It is this idea that givesthe differential equation a wide range of applications in the sciences, inthe business and social sciences.Many of the applications involve a rate ofchange of some quantity with respect to time.Thus ,if the rate of change of y with respect to time , t is proportional toy, thendydt ∝ y
⇒ dydt = ky
The constant k is referred to as the constant of proportionality.Consider a general solution to a differential equation
y = cekt
If k is positive , the function represents EXPONENTIAL GROWTH
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
If k is negative, it represents EXPONENTIAL DECAY If t=0 ,y=cHere c is referred to as the initial value (value of y at time t=0)
SOME EXAMPLES AND EXERCISES
1 In a certain type of chemical reaction, the rate at which an oldsubstance (initial amount, a) is converted into a new substance isproportional to both the amount
2 The rate of change in temperature T of a small object placed in alarge body of water with a temperature of 32◦C is proportional tothe difference between the temperature of the object and thetemperature of the water. Find the differential equation thatrepresents this function and find its general solution.
3 Newtons Law of Cooling states that the rate of decrease oftemperature of a hot body is proportional to its excess temperatureover that of the surroundings.( dθ
dt ∝ (θ − θ0)) where θ0 is thetemperature at time t.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
SOLVED EXAMPLES
Example 1
A second order rate chemical reaction is governed by the differentialequation dx
dt = k(5− x)2,where x is the change in concentration at timet. x is initially zero and is found to have the value x=1 when t=10s. Findthe value of the reaction rate constant k and the values of x when t=20sand t=100s
Solndxdt = k(5− x)2∫
dx(5−x)2 = k
∫dt = kt∫
dx(5−x)2 =?
Let u=5-x, du=-dx⇒∫
dx(5−x)2 = −
∫duu2 = −
∫u−2du = − u−1
−1 = 1u + c
= 15−x + c
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
⇒ kt = 15−x + c
when t=0,x=0 ⇒ c = − 15
⇒ kt = 15−x −
15
when t=10, x=110k = 1
4 −15 = 1
20
⇒ k = 1200
⇒ 1200 t = 1
5−x −15
when t=20, we have20200 = 1
5−x −15
15−x = 20
200 + 15 = 20
200 + 40200 = 60
200
5− x = 20060 = 10
3
x = 5− 103 = 5
3when t=100, we have100200 = 1
5−x −15
15−x = 1
2 + 15 = 7
10 ,⇒ 5− x = 107 ,⇒ x = 5− 10
7 = 257
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
CHEMICAL RATE EQUATIONS
ORDER OF A REACTION
The order of a chemical reaction is the sum of the powers of theconcentration terms that occur in the differential form of the Rateequation.Example:dxdt = k(a− x)0 ZERO ORDER REACTIONdxdt = k(a− x) FIRST ORDER REACTIONdxdt = k(a− x)2 SECOND ORDER REACTIONdxdt = k(a− x)3 THIRD ORDER REACTION
where a is initial concentration and x is the decrease in concentration inthe chemical reaction.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
FIRST ORDER REACTION
A reaction in which the rate depends on two concentration terms inwhich one species is present in a very high concentration relative to theother such that its concentration considered to be constant during thecourse of the reaction results in a first order rate equationdxdt ∝ (a− x)
⇒ dxdt = k(a− x), where a is the initial concentration and x is the
decrease in concentration.
Example
1. CH3CO − O − C2H5 + H2O CH3COOH + C2H5OHdxdt = [Ester ][H2O]
2.C (CH3)3 · OH → C4H8 + H2O
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solutiondxdt = k(a− x)∫
dxa−x = k
∫dt
− ln(a− x) = kdt + c
At t=0, x=0 ⇒ c = − ln a
⇒ − ln(a− x) = kt − ln a
⇒ ln aa−x = kt −→ Eqn1
aa−x = ekt −→ Eqn1b
SECOND ORDER REACTIONS
A + B −→ CA and B are Reactants , C is the product.If x is the decrease in concentration of A at time t, and a and b are theinitial concentrations of A and B, thendxdt = k(a− x)(b − x)For special cases in which a and b are equimolar amounts.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
dxdt = k(a− x)2
Thus∫
dx(a−x)2 = k
∫dt
⇒ (a− x)−1 = kt + CAt t=0, x=0, C= 1
a
⇒ 1a−x = kt + 1
a
kt = 1a−x −
1a
kt = xa(a−x) −→ Eqn2
Now let us consider the solution of the first order rate equationdxdt = k(a− x)
Solution:a
a−x = ekt
ln aa−x = kt
⇒ t = 1k ln a
a−x
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Usually ln(a− x) is plotted against tk is the slopeln a is the intercept.Note: For Radioisotope work
kt = ln No
Nwhere No−initial activityand N is the activity at time t.
HALF LIFE (t 12)
The half life (t 12) is the time taken for half or 50% reaction to occur
ORthe time taken for the concentration of the reactants to reduce to half(50%) of the initial concentration.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Examples
FIRST ORDER REACTIONkt = a
a−x ⇒ t = 1t ln a
a−x
HALF LIFEAt t 1
2, x = a
2
⇒ t 12
= 1k ln a
a− a2
= 1k ln 2a
2a−a = 1k ln 2
⇒ t 12
= ln 2k = 0.6932
kThus t 1
2does not depend on a
t 12
= t50% = 1k ln a
a− 50a100
= 1k ln a
50a100
= 1k ln 100a
50a = 1k ln 2
= 0.6932k
t90% = 1k ln a
a− 90a100
= 1k ln a
10a100
= 1k ln 100
10 = 1k ln 10
= 0.1054k
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
t80% = 1k ln a
a− 80a100
= 1k ln a
20a100
= 1k ln 100
20 = 1k ln 5
= 0.223k
For the second order reaction,kt = x
a(a−x)
Half life ,t 12
when x = a2
kt 12
=a2
a(a− a2 )
⇒ t 12
= 1ka
Example
In a certain thermolecular reaction the decrease x in concentration of asubstance R is given by dx
dt = k(a− x)3,where k is the reaction rateconstant and a is the initial concentration of R. Find the concentration attime t.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
Solutiondxdt = k(a− x)3∫
dx(a−x)3 = k
∫dt
12(a−x)2 = kt + C
At t=0, x=0 ⇒ 12(a−0)2 = k(0) + C ⇒ C = 1
2a2
Thus 12(a−x)2 = kt + 1
2a2
⇒ a2
(a−x)2 = 1 + 2a2kt
⇒ (a− x)2 = a2
1+2a2kt ⇒ a− x =√
a2
1+2a2kt
⇒The concentration of the substance R is given as(a− x) = a√
1+2a2kt
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
RADIOACTIVE DECAY
When radioactive substances decay, the number of atoms that decay in afixed time period is proportional to the number of atoms at the start ofthat period. Thus the rate of decay of a radioactive substance isproportional to the number of atoms N present at time t. If the constantof proportionality is λ the decay constant and initially are NO atomspresent, thendNdt ∝ N
⇒ dNdt = −λN∫
dNN = −λ
∫dt
lnN = −λt + C
When t=0, N =No ⇒ C = lnNo
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
⇒ lnN = −λt + lnNo
lnN − lnNo = −λtln N
No= −λt
NNo
= e−λt
N = Noe−λt
Example 1
Carbon 14 , one of the three isotopes of carbon is radioactive and decaysat a rate which is proportional to the amount present. Its half life i 5570years. If 10 grams were present originally, how much will be left after2000 years.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
SolutiondNdt ∝ N
⇒ dNdt = −λN∫
dNN = −λ
∫dt
lnN = −λt + C
When t=0, N =No ⇒ C = lnNo
⇒ lnN = −λt + lnNo
lnN − lnNo = −λtln N
No= −λt
NNo
= e−λt
N = Noe−λt
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS
At half life, N = No
2 , t 12
= 5570
⇒ No
2 = Noe−λ(5570)
12 = e−λ(5570)
−5570λ = ln 12
λ = − 15570 ln 1
2 = 0.000124
⇒ N = Noe−0.000124t
At t=2000 years ,No = 10grams
N = 10e−0.000124(2000)
= 7.8036 grams ' 7.8 grams.
Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]
DIFFERENTIAL EQUATIONS