1:
FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS
1.1: Introduction to Differential Equations
1.1.1: Basic Definition and Terminology
Definition:
An Ordinary Differential Equation (ODE) is an equation that contains one or several
derivatives of an unknown function.
dy
Example: 1.
= sin x + 3
dx
2. y`` + 2y` - 6y = ex
dy
d 3x
d2y
3.
+
2
- y = 0.
3
2
dx
dx
dy
Notation: F(x, y, y`, y``, … ) = 0.
Standard Notations:
If y (x) is a function of x , then the first order differential equation can be written as
dy
or y ' ( x) or y '
dx
Similarly the second order differential equation can be written as
d2y
or y ' ' ( x) or y ' '
dx 2
and in general for n th differential equation we have
dny
or y ( n ) ( x) or y (n )
n
dx
The order of an ordinary differential equation is the order of the highest derivative that
appears in the differential equation.
Example: 1.
dy
+ y tan x = sin 2x.
dx
dy
d2y
2. x2
- 4x
+ 6y = x-1.
2
dx
dx
3. y`` - 4y` + 4y = 5x2 + e-x.
4. y``` - 3y`` + 3y – y = 0.
(first order)
(second order)
(second order)
(third order)
1.1.2 : How to form ODE.
A differential equation could be formed by eliminating an arbitrary constant from a given
function.
Example 1. Form ODE from the function y = Ax + x2. (A constant)
Solution:
y = Ax + x2 … (i)
y`= A + 2x … (ii) → x(ii) : xy` = Ax + 2x2
(i) : y = Ax + x2
______________________ _
xy` - y = x2 (iii)
xy`- y = x2 . This is a first order differential equation which
derived from y = Ax + x2.
A
Example 2. Form ODE from the function y = x2 +
.
x
Solution:
A
y = x2 +
. Multiply with x, then
x
yx = x3 + A. Differenciate with respect to x,
dy
→ y+x
= 3x2 is the first order ODE.
dx
Example 3. Form ODE from the function: y = Ax2 + Bx5.
Solution:
y = Ax2 + Bx5 …. (i)
y`= 2Ax + 5Bx4…. (ii)
y``= 2A + 20Bx3…. (iii)
x(ii): xy`= 2Ax2 + 5Bx5
2(i) : 2y = 2Ax2 + 2Bx5
__________________________
xy`-2y = 3Bx …… (iv)
5
x(iii): xy`` = 2Ax + 20 Bx4
(ii):
y` = 2Ax + 5Bx4
________________________________ _
xy``- y`= 15 Bx4 ….. (v)
x(v): x2y``- xy` = 15Bx5
5(iv): 5xy` - 10y = 15Bx5
______________________________ _
x2y``- 6 xy`+ 10y = 0 (second order ODE )
Example 4. Form ODE from the function y = Aex + Be-2x
Solution:
y = Aex + Be-2x …. (i)
e (i): ye2x = Ae3x + B. …. (ii)
2x
Differentiating (ii): y`e2x + 2e2xy = 3Ae3x ….(iii)
Differentiating (iii): y``e2x + 2e2xy`+ 2e2xy`+4e2xy = 9Ae3x.
Or:
3(iii):
y``e2x + 4e2x y` + 4e2xy = 9Ae3x …. (iv)
3y`e2x + 6e2xy = 9Ae3x
______________________________________________ _
y``e2x + y`e2x - 2e2xy = 0
e2x(y``+ y` - 2y) = 0.
Thus the solution is:
But e2x ≠ 0
y`` + y` - 2y = 0
1.1.3: Concepts of Solution and Initial Value Problem
(A) Solution of a Differential Equation.
Definition: If y = F(x) is the solution of an ODE, hence a function F(x) satisfies the
given differential equation.
dy
d2y
Example. 5. Given
+
- 6y = 0. Show that:
2
dx
dx
(a)
y = e2x is the solution.
(b)
y = 5e2x + 4e-3x is the solution.
(c)
y = xe2x is not the solution.
Solution:
dy
d2y
= 2e2x …(ii) and
= 4e2x …(iii)
dx
dx 2
Substitute (i), (ii) dan (iii) into the given diff. eq. hence
(a) y = e2x…(i) thus
dy
d2y
+
- 6y = 4e2x + 2e2x – 6e2x = 0.
2
dx
dx
It is shown that y = e2x is the solution.
(b) y = 5e2x + 4e-3x
dy
= 10e2x – 12e-3x
dx
d2y
= 20e2x + 36e-3x
dx 2
dy
d2y
→
+
- 6y = 20e2x + 36e-3x + 10e2x – 12e-3x
2
dx
dx
-30e2x – 24e-3x
= 0
2x
-3x
y = 5e + 4e is the solution.
(c)
→
y = xe2x
y` = 2xe2x + e2x
y``= 2e2x + 4xe2x + 2e2x = 4xe2x + 4e2x.
y``+ y` - 6y = 4xe2x + 4e2x + 2xe2x + e2x – 6e2x
= 5e2x ≠ 0.
y = xe2x is not the solution.
Example 6.
Find the value of m so that y = emx is the solution of the diffrential equation
2y`` + 5y` - 3y = 0.
Solution:
Given y = emx …..(i), thus y`= memx …(ii) and y``= m2emx …(iii)
Substitute (i), (ii) and (iii) into the ODE, hence
2y``+ 5y` - 3y =0
2m2emx + 5memx – 3emx =0
emx(2m2 + 5m – 3) = 0. But emx ≠ 0 hence,
2m2 + 5m – 3 = 0
(2m -1)(m + 3) = 0
m = { ½ , -3}.
(B) General & Particular Solution (Initial Value Problem).
Definition:
(i) General Solution
Solution obtained from integrating equations are called general solutions.
(ii) Particular Solution
Particular solution are the solutions obtained by assigning specific values to the arbitrary
constants in the general solutions.
Example 7. Show that y = Aex + (x + 2)e2x is the general solution of the differential
dy
equation
- y = (x + 3)e2x, and hence determine the value of A given that y = 4 when
dx
x = 0.
Solution:
y = Aex + (x + 2)e2x
dy
= Aex + 2(x + 2)e2x + e2x
dx
= Aex + (2x + 5)e2x
dy
→
- y = Aex +(2x + 5)e2x – Aex – (x + 2)e2x
dx
= (2x + 5 – x – 2)e2x
= (x + 3)e2x. (shown)
Given that y = 4 when x = 0
→ y = Aex + (x + 2)e2x
4 = Ae0 + (0 + 2)e0
4=A+2
→ A=2
Particular solution: y = 2ex + (x + 2)e2x
The particular solution could be obtained by substituting the given condition (y = 4 when
x = 0). The conditions are called the initial condition of the differential equation.
Definition:
(i) Initial Value Problem (IVP) is a differential equation with initial conditions.
(Ex. y = 1 and y`= 2 when x = 0)
(ii) Boundary Value Problem (BVP) is a diff. equation with boundary conditions.
(Ex. y = 0 when x = 0 and y`= 2 when x = 1)
Example 8.
Akos3 x  B sin 3x
dy
d2y
is the general solution for x 2 + 2
+ 9xy = 0.
x
dx
dx
And hence obtain the particular solution with condition y( ) = -3 and y`( ) = 0.
Show that y =
Solution:
The conditions above are an initial condition (IVP)
y = -3 and y`= 0 when x =
Given:
yx = A kos3x + B sin 3x … (i)
dy
x
+ y = -3A sin3x + 3B kos3x … (ii)
dx
dy
d 2 y dy
+
+
= -9A kos3x – 9B sin3x.
2
dx
dx
dx
dy
d2y
x 2 +2
= -9(A kos3x + B sin3x) … (iii)
dx
dx
Substitute (i) into (iii), thus:
x
x
dy
d2y
+2
+ 9xy = 0. (shown)
2
dx
dx
Substitute y( ) = -3 into (i) → -3 = -A or A = 3
y`( ) = 0 into (ii) → y = -3B
-3 = -3B or B = 1.
3kos3 x  sin 3 x
The particular solution: y =
.
x
Example 9.
B
is the general solution for x2y`` + xy` - 9y = 0 and hence obtain
x3
the particular solution with conditions y(2) = 1 and y`(1) = 0.
Show that y = Ax3 +
Solution:
The condition above are a boundary condition (BVP), y(2) = 1 and y`(1) = 0.
B
or x3y = Ax6 + B … (i).
x3
Differentiating (i), thus 3x2y + x3y`= 6Ax5
→ xy` = 6Ax3 – 3y … (ii).
y = Ax3 +
Differentiating (ii), thus xy``+ y`= 18Ax2 – 3y`
→ xy``= 18Ax2 – 4y` …(iii)
Substitute (ii) and (iii) into given diff. equation,
x2y``+ xy`- 9y = 18Ax3- 4y`x + xy`- 9y
= 18Ax3 -3(6Ax3- 3y) – 9y
= 18Ax3 – 18Ax3 + 9y – 9y
=0
B
Thus: y = Ax3 + 3 is the general solution.
x
Substituting y(2) = 1 or y = 1 when x = 2
B
into diff. equation y = Ax3 + 3 we get
x
1
B or 8 = 64A + B…(iv)
8
Substituting y`(1) = 0 or y`= 0 when x = 1
into xy`= 6Ax3 – 3y we get
B
xy`= 6Ax3 – 3(Ax3 + 3 )
x
3B
xy` = 3Ax3 - 3
x
0= 3A – 3B → A = B … (v)
1 = 8A +
From simultaneous equation (iv) and (v), thus
8
64A + A = 8 → A = B =
65
8 3
1
Particular equation: y =
(x + 3 ).
65
x
1.2: First Order Ordinary Differential Equation(ODE)
General Form:
dy
= f(x,y)
dx
dy
= 2y + sin x.
dx
dy
x 2  (1  x) y
b)
=
.
dx
2x
Example: a)
There are four types of a first order ODE,
i)
Separable differential equation.
ii)
Homogeneous differential equation.
iii)
Linear differential equation.
iv)
Exact differential equation.
1.2.1: Separable Differential Equation.
The differential equation:
y` = f(x,y)
is said to be separable if the equation can be written as the product of a function of x, u(x)
and the function of y, v(y).
The equation can be written in the form
dy
dy
= u(x).v(y) or
= u(x).dx
dy
v( y )
hence, integrate both sides:
∫
dy
= ∫ u(x) dx.
v( y )
Example 10. Solve the equation: (x + 2)
Solution:
dy
= y
dx
dx
dy
∫
= ∫
x2
y
ln|y| = ln|x+2| + C
y
ln|
| = ec = A
x2
y = A(x+2).
(x + 2)
dy
= y.
dx
Example 11. Solve the equation: ex
dy
+ xy2 = 0.
dx
Solution:
dy
+ xy2 = 0.
dx
dy
∫ 2 = - ∫ xe-xdx.
y
d ( x)
1
= -[x ∫e-xdx - ∫{e-xdx}
dx.
dx
y
1
= -xe-x -∫-e-xdx
y
1
= -xe-x – e-x + C.
y
1
= -(x+1) e-x + C.
y
ex
Example 12.
Solve the following differential equation: x2y dx + (x + 1) dy = 0 which satisfied
condition y = 2 when x = 0.
Solution:
x2y dx + (x + 1) dy = 0
dy
x2
=
dx.
y
x 1
1
dy
= {(x – 1) +
}dx.
x 1
y
dx
dy
-∫
= ∫(x – 1)dx + ∫
x 1
y
x2
-ln|y| =
- x + ln|x + 1| + C.
2
ln|y(x + 1)| = x – ½ x2 – C.
y(x + 1) = ex-1/2 x
2
-C
2
y(x + 1) = A.ex-1/2 x , where A = e-C
y = 2 when x = 0, thus: 2 = A.
The solution is:
y=
2
2
. ex- ½ x
x 1
Substitution Method.
Example 13.
Solve the equation:
Solution
:
dy
x  y 1
=
which satisfied the condition y(1) = 1.
dx
x y 5
Subsitute z = x + y
dz
dy
 1+
thus
dx
dx
dz
z 1
→
-1 =
dx
z 5
dz
z 1
=
+1=
dx
z 5
dy
dz
=
-1
dx
dx
2z  6
2( z  3)
=
z5
z5
z 5
dz = 2 dx.
z3
2
∫(1 +
) dz = ∫2 dx.
z 3
z + 2ln|z+3| = 2x + C.
2ln|z+3| = 2x – x – y + C
(z + 3)2 = A.ex-y, where A = eC.
y(1) = 1 → (1+1+3)2 = A.e1-1
25
The solution is: (x + y + 3)2 = 25 ex-y .
Example 14. Solve the equation:
x
dy
+ y = 2x((1 + x2y2).
dx
Solution: Substitute z = xy, hence
dz
dy
x  y
dx
dx
→
dz
= 2x(1 + z2)
dx
dz
∫
= ∫ 2xdx.
1 z2
tan-1 z = x2 + C.
z = tan(x2 + C)
= A
xy = tan(x2 + C).
tan( x 2  C )
y=
x
1.2.2:Homogeneous Equations
Consider the differential equation
dy
= f(x, y).
dx
If: f(λx, λy) = f(x, y) for each   , hence
dy
= f(x, y) is called a homogeneous
dx
equation.
dy
xy
= 2
= f(x, y)
dx
x  y2
Example: i).
(x)(y )
2 ( xy)
=
( x ) 2  ( y ) 2
2 ( x 2  y 2 )
xy
= 2
= f(x, y) [homogeneous].
x  y2
f(λx, λy) =
ii).
dy
= x – y = f(x, y).
dx
f(λx, λy) = λx – λy = λ(x – y) ≠ f(x, y).
f(x, y) non-homogeneous.
The method of solving a homogenous diff. equation is by using the following
substitution.
y = x.v, hence
dy
dv
=x
+v
dx
dx
Example 15. Solve the differential equation
dy
xy
= 2
with condition y(0) = 2.
dx
x  y2
Solution: By using substitution y = xv and
dy
dv
= x + v.
dx
dx
Thus: x
x
x( xv)
dv
v
+v= 2
=
2
dx
x  ( xv)
1 v2
dv
v
v  v(1  v 2 )
v3
=
v
=
=
dx 1  v 2
1 v2
1 v2
dx
1 v2
) dv = - ∫
dx.
3
x
v
1
+ ln |v| = -ln|x| + C.
 2v 2
1
y
ln |xv| =
+ C.
[v = ]
2
x
2v 2 2
x /2y
y = A.e
, where A = eC
∫(
Then y(0) = 2 , hence A = 2.
The solution is: y = 2ex
2
/2y
2
Example 16. Solve the differential equation
dy
2x  y
=
with condition y(3) = 1.
dx
x  2y
Solution:
2 x   y
 (2 x  y )
2x  y
=
=
= f(x, y).
x  2y  ( x  2 y )
x  2y
dy
dv
 x  v , hence
Substitute y = xv and
dx
dx
dv
2 x  xv
2v
x
+v=
=
.
x  2 xv
dx
1  2v
dv
2v
 2(v 2  1)
x
=
-v =
.
dx
1  2v
2v  1
f(λx, λy) =
2v  1
dx
)dv = ∫-2
2
x
v 1
dx
1
3
∫{
+
}dv = -∫ 2 .
x
2(v  1)
2(v  1)
∫(
1
3
ln|v + 1| + ln|v – 1| = -2ln|x| + C
2
2
ln|v + 1| + 3ln|v – 1| = -4ln|x| + 2C
(v + 1)(v – 1)3.x4 = A , where A = e2C
yx yx 3 4
(
)(
) .x
=A
x
x
(y + x)(y – x)3 = A
The condition y(3) = 1 → A = -32.
The solution is: (y + x)(y – x)3 + 32 = 0.
1.2.3: Linear Equations.
dy
+ b(x).y = c(x).
dx
dy b( x)
c( x)
+
.y =
dx a ( x)
a( x)
dy
or:
+ p(x).y = q(x)
dx
b( x )
c( x)
where p(x) =
and q(x) =
This is the general form of a linear differential
a( x)
a( x)
equations.
Note:
a(x)
The Method of Solution.
ii)
dy
+ p(x).y = q(x)
dx
Determine p(x) and evaluate : ∫ p(x) dx.
iii)
Obtain the integrating factor : u(x) = e∫ p(x)dx.
iv)
u(x)
i)
v)
vi)
vii)
Write to the general form :
dy
+ u(x).p(x).y = u(x).q(x).
dx
d
Write
{u(x).y} = u(x).q(x).
dx
∫ d(u(x).y = ∫ u(x).q(x)dx.
u(x).y = ∫ u(x).q(x)dx.
dy
+ y = x3
dx
dy
ii).
- y = 2ex
dx
dy
iii). (1 + x2)
- xy = x(1 + x2)
dx
Example: i). x
Solution
dy
i). x
+ y = x3
dx
dy
y
+
= x2
dx
x
1
1
p(x) =
→ ∫p(x)dx = ∫ dx = ln x.
x
x
Integrating factor: u(x) = e∫p(x)dx = eln x = x.
y.x = ∫x.x2 dx = ∫x3dx
1
= x4 + C
4
1
C
→ y = x3 +
.
4
x
dy
- y = 2ex.
dx
p(x) = -1 → ∫p(x) dx = ∫(-1) dx = -x.
ii).
Integrating factor: u(x) = e∫p(x) dx = e-x.
e-x.y = ∫e-x.2ex dx = 2x + C
→
y = 2xex + Cex.
dy
- xy = x(1 + x2)
dx
dy
x
-(
).y = x
dx
1 x2
x
∫p(x)dx = ∫-(
)dx = ln(1 + x2)-x/2
2
1 x )
iii): (1 + x2)
2
1 / 2
u(x) = e∫p(x)dx = eln(1+x )
= (1+x2)-1/2
x
(1+x2)-1/2.y = ∫(
)dx. Substitute z = (1+x2)
2 1/ 2
(1  x )
x
hence ∫(
)dx = (1 + x2)1/2 + C
(1  x 2 )1 / 2
(1+x2)-1/2.y = (1 + x2)1/2 + C.
→ y = (1 + x2) + C(1 + x2)1/2
1.2.4: Exact Equations.
General form:
M(x,y) dx + N(x,y) dy = 0.
Condition of an Exact Equation:
Example:
M
N

y
x
i) (2x + 3y2) dx + (6xy + 2y) dy = 0
ii) (3x2y + ey) dx + (x3 + xey – 2y) dy = 0
iii)
(2x + y – kos y) dx + (4y + x + sin x) dy = 0.
The method of solution.
a)
M dx + N dy = 0. Test for exactness:
M
N

y
x
u
= M …….. (i)
x
u
= N ………(ii)
y
c) Integrate with renpect to x: ∫ du = ∫ M dx
u = ∫ Mdx + Q(y) …..(iii)
d) Differentiate (iii) with respect to y.
b)
Write
e) Equate: u(x,y) = A.
Example: Solve the following differential equation.
(6x2 – 10 xy + 3y2) dx + (6xy – 5x2 – 3y2) dy = 0
Solution
Step 1:
M ( x, y )  6x2 – 10 xy + 3y2
N ( x, y )  6xy – 5x2 – 3y2
Step 2:
M  10 x  6 y
N  6 y  10 x
So M  N  6 y  10 x (exact)
y
x
y
x
Step 3:
 du   (6 x  10 xy  3 y )dx
F ( x, y)  2 x  5 x y  3xy  g ( y )
2
2
3
2
2
Step 4:
F  5 x  6 xy  g ( y )
2
y
Step 5:
F  N ( x, y )
 5 x  6 xy  g ( y )  6 xy  5 x  3 y
So we get g ( y )  3 y , then integrate with respect to y
g ( y)   y
Then
y
2
2
2
2
3
Final step:
F ( x, y)  2 x  5 x y  3xy  y
3
2
2
3
Exercises
1. Solve the differential equations:
dy
i)
+ 3y = e2x
dx
dy
ii)
+ y = x2
dx
dy
iii) sin x
+ 2y kos x = kos x
dx
dy
iv) sin x
- y kos x = cot x.
dx
[y=
1 2x
e + Ce-3x]
5
[ y = x2- 2x + 2 + Ce-x]
[ysin2x = A-
1
kos2x]
4
1
[y = - kosek x+Csin x]
2
2. Show that these equations is exact and solve.
1
y
i) (y3 ) dy + 2 dx = 0
x
x
ii) (3x2 – y sin xy) dx – x sin xy = 0
iii) (2x + 3 kos y) dx + (2y – 3x sin y) dy = 0.
A summary of strategies for identifying and solving various kinds of first order
differential equation is displayed in table given below.
Table : Summary of strategies for solving first-order differential equation
Form of Equation
Method
Solution
dy g ( x)

dx f ( y )
Separate the variables.
 f ( y)dy   g ( x)dx
dy
 f  x, y   f   x ,  y 
dx
Homogeneous- use a change of
 f (v)dv   x dx,
(check for homogeneous)
dy
 p( x) y  q ( x)
dx
variable, y  vx and
dy
dv
 x v
dx
dx
Use the integrating factor
  e
Pdx
M ( x, y )dx  N ( x, y )dy  0,
Exact- use partial integration to
where
find f, where
M N

y
x
f
f
 M and
N
x
y
(check for exactness)
1
v y/x
y    Q  dx
y
1

 Q dx
f ( x, y )  C
1.3: Applications of First Order Differential Equations
Modeling with differential equations is the process of writing a differential equation to
describe a physical situation. We will look at three different situations which involve
applications of first order differential equations, namely Newton’s Law of Cooling,
population growth and mixing problems.
1.3.1: Newton;s Law of Cooling
Newton's Law of Cooling is used to model the temperature change of an object of some
temperature placed in an environment of a different temperature. The law states that:
dT
  k T  Ts  ….(i)
dt
where T is the temperature of the object at time t, Ts is the temperature of the surrounding
environment (constant) and k is a constant of proportionality. This law says that the rate
of change of temperature is proportional to the difference between the temperature of
the object and that of the surrounding environment in which it is placed. Examples of
cooling problems include a cup of coffee set down in a room, a hot pizza that is taken out
of the oven and allowed to cool on the kitchen counter, etc.
In order to get the equation (i) to something that we can use, we must solve the
differential equation. The steps are given below.
1. Separate the variables : Get all the T’'s on one side and all the t's on the other side.
dT
 k dt
The constants can be on either side:
T  Ts
2. Integrate both sides
3. Solve for T
ln T  Ts   C  kt
:
:
T  Ae  kt  Ts , where A  e C
We now have a useful equation. When you are working with Newton's Law of Cooling,
remember that t is the variable. The other letters, M, k, A, are all constants. In order to
find the temperature of the object at a given time, all of the constants must first have
numerical values.
Let’s look at examples of Newton’s Law of Cooling, population growth and Mixing
Problems.
Example 1: Newton’s Law of cooling
According to Newton’s Law of cooling, the rte at which body cools is given by the
dT
  k T  Ts  ,
equation
dt
where Ts is the temperature of surrounding medium, k is a constant and t is the time in
minutes. If the body cools from 80°C to 50°C in 20 minutes which the surrounding
temperature of 10 °C, how long does it need for the body to cool from 80°C to 30°C and
also time taken for the body to cool from 50°C to 30°C.
Exercise:
A boiling (100°C) solution is set on a table where room temperature is assumed to be
constant 20°C. Find the solution cooled to 60°C after five minutes.
a. Find a formula for the temperature (T) of the solution, t minutes after it is placed on
the table.
b. Determine how long it will take for the solution to cool to 22°C
Answer: a. T  20  80e 0.13863t
b. t=26.6 minutes
Example 2: Time of Death
Suppose that a corpse was discovered in a motel room at midnight and its temperature
was
. The temperature of the room is kept constant at
. Two hours later the
temperature of the corpse dropped to
. Find the time of death. Normal body
temperature is
(assuming the dead person was not sick!).
Solution: The death happened around 7:28 P.M.
Example 3: Crime Scene
A detective is called to the scene of a crime where a dead body has just been found. She
arrives on the scene at 10:23 pm and begins her investigation. Immediately, the
temperature of the body is taken and is found to be 80o F. The detective checks the
programmable thermostat and finds that the room has been kept at a constant 68o F for
the past 3 days.
After evidence from the crime scene is collected, the temperature of the body is taken
once more and found to be 78.5o F. This last temperature reading was taken exactly one
hour after the first one. The next day the detective is asked by another investigator,
“What time did our victim die?” Assuming that the victim’s body temperature was
normal (98.6o F) prior to death, what is her answer to this question? Newton's Law of
Cooling can be used to determine a victim's time of death.
Answer: 3.23 pm
1.3.2:Population Growth
Differential equations whose solutions involve exponential growth or decay are
discussed. Everyday real-world problems involving these models are also introduced.
Consider the differential equation
dP
kP
dt
Using the method of separation of variables we find
dP
 k dt
P
Integrating both sides to obtain
ln P  kt  C
Or
P  Ae kt
Note that P0  P0
P  P0 e kt
Thus, We say that the solution represents a growth model when k > 0 and a decay model
when k < 0:
Applications for Growth/Decay Models
Population Dynamics
If P(t) is the population of a species at time t then by the \conservation of population" law
the rate of change of the population is the difference of the rate of population increase,
due for example to birth, and the rate of population decrease, due for example to deaths.
In mathematical model, we have
dP
 rb P  rd P  kP
dt
Assuming that no migration exists. Thus, the population at time t is given by
Pt   P0 e kt
Example 4
Around thousands citizens are living in a small country and this amount has been
changed to 450 thousands within 2 years. The population grows at rate that is
proportional to the number of the citizens present at the time. Assume that the 50
thousands is the initially, determine
a) The equation of related the size of population to the time
b) How many citizen of the small country in this population after 5 years
.Answer: P=12,149,254 people
Example 5
Suppose that the population of a colony of bacteria increases exponentially. At the start
of an experiment, there are 6,000 bacteria, and one hour later, the population has
increased to 6,400. How long will it take for the population to reach 10,000? Round your
answer to the nearest hour.
Answer: t=8 hours
1.3.4: Mixing Problems
The mixing problems we discuss here will involve a tank into which certain mixture will
be added at a certain input rate and the mixture will leave the system at a certain output
rate. The problem is to determine the amount of salt in the tank at any time. We will use y
= y(t) to denote the amount of mixture in the tank at any given time t. The differential
equation involved is from the following natural relationship:
Rate of change of y(t) = Rate at which y(t) enters the tank - Rate at which y(t) leave
the tank
or
dy
 input rate – output rate
dt
where,
Rate at which
y(t) enters the
tank
= ( flow rate of liquid
entering )
x
( concentration of substance in
liquid
entering )
Rate at which
y(t) leave the
tank
= ( flow rate of liquid
leaving )
x
( the concentration of substance in
liquid
leaving)
In this mixing problems, we will discuss a typical mixing problem which deals with the
amount of salt in a mixing tank.
Example 6:
Consider a tank with volume 100 liters containing salt solution. Suppose a solution with
2kg per litter of salt flows into the tank at a rate of 5 liters per minute. The solution in the
tank is well mixed. Solution flows out of tank at a rate of 5 liters per minute. If initially
there is 20kg of salt in the tank, determine the amount of salt will be in the tank at any
time?
Example 7:
Initially 50kg of salt is dissolved in a large tank holding 300 liters of water. A brine
solution is pumped into the tank at a rate of 3 liters per minute, and the well-stirred
solution is then pumped out at the same rate. If the concentration of the solution
entering is 2 kg per liter, determine the amount of salt in the tank at any time.
How much salt is present after 50 min?
Example 8:
Initially a water tower contains 1 million liters of pure water. Two valves are then
opened, one allowing a solution of water and fluoride with a concentration of 0.1kg of
fluoride per liters of water to flow into the system at a rate of 80 liters/minute, and the
other valve allows the solution in the tank to be drained at 80 liters/minute. Assuming
that the solution is mixed constantly so that we have a homogeneous fluid in the tank,
(a) Find an expression for the amount (in kg) of fluoride in the water tower after t
minutes
(b) Find how long it takes for the concentration to come to a level of 0.05 kg/l
Answer:
(a) y(t )  1000000  1000000e0.00008t
(b) t  8664.34 minutes
Example 9:
You started with 10 g of radioactive Q, but after 6 days of decay there were only 3 g left.
(a) Find a formula for the amount of Q present after t days.
(b) What is the half–life of Q?
Answer:
(a) M (t )  10e0.2007t
(b) t  3.45 days
Example 10:
Suppose medicine M has an absorption (decay) constant of –0.17 (determined
experimentally), and that the lowest concentration of M that is "effective" is 0.3 mg/l
(milligrams of M per liter of blood). If a patient who has 8 liters of blood is injected with
20 mg of M, how long will the M be effective?
Answer:
(a) M (t )  2.5e0.17
(b) t  12.47 hours