CHAPTER 2. FIRST ORDER ODEs
2.3 Linear Equations
Calistus Ngonghala
University of Florida
CNN
2.3 Linear Equations
Outline
1
Introduction
2
Identifying first order linear equations
3
Solving first order linear equations
4
Initial Value Problems
Introduction
In this chapter we are learning methods to solve ODEs.
Introduction
In this chapter we are learning methods to solve ODEs.
The main principle involved in solving ODEs is to:
Introduction
In this chapter we are learning methods to solve ODEs.
The main principle involved in solving ODEs is to:
recognize the class to which the ODE belongs
Introduction
In this chapter we are learning methods to solve ODEs.
The main principle involved in solving ODEs is to:
recognize the class to which the ODE belongs
learn and use the methods that allow us to solve this class of
ODE problems
Introduction
In this chapter we are learning methods to solve ODEs.
The main principle involved in solving ODEs is to:
recognize the class to which the ODE belongs
learn and use the methods that allow us to solve this class of
ODE problems
In this Chapter we only work with first order ODEs.
Introduction
In this chapter we are learning methods to solve ODEs.
The main principle involved in solving ODEs is to:
recognize the class to which the ODE belongs
learn and use the methods that allow us to solve this class of
ODE problems
In this Chapter we only work with first order ODEs.
We will consider first order linear equations.
Identifying first order linear equations
Like separable ODEs, linear can be solved easily.
Identifying first order linear equations
Like separable ODEs, linear can be solved easily.
Definition
A first order linear ODE is an equation of the form:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
(1)
where a1 (x) 6= 0, a0 (x), and b(x) depend only on x but not on y.
Identifying first order linear equations
Like separable ODEs, linear can be solved easily.
Definition
A first order linear ODE is an equation of the form:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
(1)
where a1 (x) 6= 0, a0 (x), and b(x) depend only on x but not on y.
At times the ODE will not be written in the form (1) so that
we can see right away that it is linear.
Identifying first order linear equations
Like separable ODEs, linear can be solved easily.
Definition
A first order linear ODE is an equation of the form:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
(1)
where a1 (x) 6= 0, a0 (x), and b(x) depend only on x but not on y.
At times the ODE will not be written in the form (1) so that
we can see right away that it is linear.
We may need to rewrite it in the form above or use methods
from Chapter 1 to identify a linear equation.
Identifying first order linear equations
Like separable ODEs, linear can be solved easily.
Definition
A first order linear ODE is an equation of the form:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
(1)
where a1 (x) 6= 0, a0 (x), and b(x) depend only on x but not on y.
At times the ODE will not be written in the form (1) so that
we can see right away that it is linear.
We may need to rewrite it in the form above or use methods
from Chapter 1 to identify a linear equation.
Some ODEs are both separable and linear
Identifying linear equations
Example
Determine whether the given first order ODE is linear or non-linear
ex cos x + (sin x)y = x3
dy
dx
Identifying linear equations
Example
Determine whether the given first order ODE is linear or non-linear
ex cos x + (sin x)y = x3
dy
dx
Solution
First we check whether the equation is written in the form
(1). If not, we try to write the equation in this form:
Identifying linear equations
Example
Determine whether the given first order ODE is linear or non-linear
ex cos x + (sin x)y = x3
dy
dx
Solution
First we check whether the equation is written in the form
(1). If not, we try to write the equation in this form:
x3
dy
− (sin x)y = ex cos x
dx
Hence, here
a1 (x) = x3
a0 (x) = − sin x
b(x) = ex cos x
Identifying linear equations
Example
Determine whether the given first order ODE is linear or non-linear
ex cos x + (sin x)y = x3
dy
dx
Solution
First we check whether the equation is written in the form
(1). If not, we try to write the equation in this form:
x3
dy
− (sin x)y = ex cos x
dx
Hence, here
a1 (x) = x3
a0 (x) = − sin x
We conclude the equation is linear.
b(x) = ex cos x
Example 1 – Problem #4/p.54
Example
Determine whether the given equation is separable, linear, both or
dy
+ y ln t
neither.
3t = et
dt
Solution
Example 1 – Problem #4/p.54
Example
Determine whether the given equation is separable, linear, both or
dy
+ y ln t
neither.
3t = et
dt
Solution
First, we express the ODE in the form (1): pause
et
dy
+ (ln t)y = 3t
dt
Example 1 – Problem #4/p.54
Example
Determine whether the given equation is separable, linear, both or
dy
+ y ln t
neither.
3t = et
dt
Solution
First, we express the ODE in the form (1): pause
et
dy
+ (ln t)y = 3t
dt
Hence,
a1 (t) = et
a0 (t) = ln t
b(t) = 3t
Example 1 – Problem #4/p.54
Example
Determine whether the given equation is separable, linear, both or
dy
+ y ln t
neither.
3t = et
dt
Solution
First, we express the ODE in the form (1): pause
et
dy
+ (ln t)y = 3t
dt
Hence,
a1 (t) = et
a0 (t) = ln t
We conclude the equation is linear.
b(t) = 3t
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Then check whether the right-hand side is of the form:
f (t, y) = g(t)p(y)
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Then check whether the right-hand side is of the form:
f (t, y) = g(t)p(y)
Solve the ODE for y 0 :
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Then check whether the right-hand side is of the form:
f (t, y) = g(t)p(y)
Solve the ODE for y 0 :
dy
3t − ln ty
=
dt
et
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Then check whether the right-hand side is of the form:
f (t, y) = g(t)p(y)
Solve the ODE for y 0 :
dy
3t − ln ty
=
dt
et
The numerator of the right-hand side cannot be factored as a
function of t times a function of y.
Example 1 – Problem #4/p.54
Solution
To check for separability, we write the equation in the form
dy
= f (t, y)
dt
Then check whether the right-hand side is of the form:
f (t, y) = g(t)p(y)
Solve the ODE for y 0 :
dy
3t − ln ty
=
dt
et
The numerator of the right-hand side cannot be factored as a
function of t times a function of y.
We conclude the equation is NOT separable.
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Dependent variable: and dependent r
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Dependent variable: and dependent r
We know that because of
dr
dθ
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Dependent variable: and dependent r
We know that because of
dr
dθ
Linearity
Express the equation in the form (1):
dr
− 3r = θ3
dθ
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Dependent variable: and dependent r
We know that because of
dr
dθ
Linearity
Express the equation in the form (1):
Hence,
a1 (θ) = 1
a0 (θ) = −3
dr
− 3r = θ3
dθ
b(θ) = θ3
Example 2 – Problem #6/p.54
Example
Determine whether the given equation is separable, linear, both or
dr
− θ3
neither.
3r =
dθ
Solution
First we identify the independt and dependent variables
Independent variable: θ
Dependent variable: and dependent r
We know that because of
dr
dθ
Linearity
Express the equation in the form (1):
Hence,
a1 (θ) = 1
a0 (θ) = −3
We conclude the equation is linear
dr
− 3r = θ3
dθ
b(θ) = θ3
Example 2 – Problem #6/p.54
Solution
To check for separability, we write the equation in the form
dr
= f (θ, r)
dθ
Example 2 – Problem #6/p.54
Solution
To check for separability, we write the equation in the form
dr
= f (θ, r)
dθ
and we can write the right-hand side as
f (θ, r) = g(θ)p(r)
Example 2 – Problem #6/p.54
Solution
To check for separability, we write the equation in the form
dr
= f (θ, r)
dθ
and we can write the right-hand side as
f (θ, r) = g(θ)p(r)
Solve the ODE for r0 :
dr
= θ3 + 3r
dθ
Example 2 – Problem #6/p.54
Solution
To check for separability, we write the equation in the form
dr
= f (θ, r)
dθ
and we can write the right-hand side as
f (θ, r) = g(θ)p(r)
Solve the ODE for r0 :
dr
= θ3 + 3r
dθ
The right-hand side cannot be factored as a function of θ
times a function of r.
Example 2 – Problem #6/p.54
Solution
To check for separability, we write the equation in the form
dr
= f (θ, r)
dθ
and we can write the right-hand side as
f (θ, r) = g(θ)p(r)
Solve the ODE for r0 :
dr
= θ3 + 3r
dθ
The right-hand side cannot be factored as a function of θ
times a function of r.
We conclude the equation is NOT separable.
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
To understand how the method emerged, we introduce the
method in steps.
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
To understand how the method emerged, we introduce the
method in steps.
Case 1. Assume a0 (x) = 0.
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
To understand how the method emerged, we introduce the
method in steps.
Case 1. Assume a0 (x) = 0. Then the equation (1) becomes:
a1 (x)
dy
= b(x)
dx
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
To understand how the method emerged, we introduce the
method in steps.
Case 1. Assume a0 (x) = 0. Then the equation (1) becomes:
a1 (x)
dy
= b(x)
dx
We may rewrite this equation as
dy
b(x)
=
dx
a1 (x)
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
After ascertaining that the equation is linear, we need a
method to solve it
To understand how the method emerged, we introduce the
method in steps.
Case 1. Assume a0 (x) = 0. Then the equation (1) becomes:
a1 (x)
dy
= b(x)
dx
We may rewrite this equation as
dy
b(x)
=
dx
a1 (x)
This equation is separable and can be solved as such.
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
a1 (x)
dy
+ a01 (x)y = b(x)
dx
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
d
(a1 (x)y) = b(x)
dx
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
d
(a1 (x)y) = b(x)
dx
Integrating both sides with respect to x, we have
Z
Z
d
(a1 (x)y) dx = b(x) dx
dx
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
d
(a1 (x)y) = b(x)
dx
Integrating both sides with respect to x, we have
Z
Z
d
(a1 (x)y) dx = b(x) dx
dx
Hence,
Z
a1 (x)y =
b(x) dx + C
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
d
(a1 (x)y) = b(x)
dx
Integrating both sides with respect to x, we have
Z
Z
d
(a1 (x)y) dx = b(x) dx
dx
Hence,
Z
a1 (x)y =
b(x) dx + C
Solving for y, we have an explicit solution:
Solving first order linear equations
Linear ODE:
a1 (x)
dy
+ a0 (x)y = b(x)
dx
Case 2. Assume a0 (x) = a01 (x). Then the equation (1) becomes:
dy
+ a01 (x)y = b(x)
dx
We may rewrite this equation as
a1 (x)
d
(a1 (x)y) = b(x)
dx
Integrating both sides with respect to x, we have
Z
Z
d
(a1 (x)y) dx = b(x) dx
dx
Hence,
Z
a1 (x)y =
b(x) dx + C
Solving for y, we have an explicit solution:
1
y=
a1 (x)
Z
b(x) dx + C
Example 3 – Not a book problem
Example
Obtain a general solution to the equation
dy
= x2 + 2x − 1 − 2xy
(x2 + 1)
dx
Example 3 – Not a book problem
Example
Obtain a general solution to the equation
dy
= x2 + 2x − 1 − 2xy
(x2 + 1)
dx
Solution
First, we recognize that the equation is linear and write it in
the linear ODE form
Example 3 – Not a book problem
Example
Obtain a general solution to the equation
dy
= x2 + 2x − 1 − 2xy
(x2 + 1)
dx
Solution
First, we recognize that the equation is linear and write it in
the linear ODE form
(x2 + 1)
dy
+ 2xy = x2 + 2x − 1
dx
Example 3 – Not a book problem
Example
Obtain a general solution to the equation
dy
= x2 + 2x − 1 − 2xy
(x2 + 1)
dx
Solution
First, we recognize that the equation is linear and write it in
the linear ODE form
(x2 + 1)
dy
+ 2xy = x2 + 2x − 1
dx
We recognize that
a1 (x) = x2 + 1
a0 (x) = 2x
We can see that a0 (x) = a01 (x).
b(x) = x2 + 2x − 1
Example 3 – Not a book problem
Example
Obtain a general solution to the equation
dy
= x2 + 2x − 1 − 2xy
(x2 + 1)
dx
Solution
First, we recognize that the equation is linear and write it in
the linear ODE form
(x2 + 1)
dy
+ 2xy = x2 + 2x − 1
dx
We recognize that
a1 (x) = x2 + 1
a0 (x) = 2x
We can see that a0 (x) = a01 (x).
b(x) = x2 + 2x − 1
Example 3 – Not a book problem
Solution
We rewrite the left-hand side as
d
[(x2 + 1)y] = x2 + 2x − 1
dx
Check!
Example 3 – Not a book problem
Solution
We rewrite the left-hand side as
d
[(x2 + 1)y] = x2 + 2x − 1
dx
Check!
Integrating both sides with respect to x, we have
Z
d
[(x2 + 1)y] dx =
dx
Z
x2 + 2x − 1 dx
Example 3 – Not a book problem
Solution
We rewrite the left-hand side as
d
[(x2 + 1)y] = x2 + 2x − 1
dx
Check!
Integrating both sides with respect to x, we have
Z
d
[(x2 + 1)y] dx =
dx
Z
x2 + 2x − 1 dx
We obtain and implicit solutition
(x2 + 1)y =
x3
+ x2 − x + C
3
Example 3 – Not a book problem
Solution
We rewrite the left-hand side as
d
[(x2 + 1)y] = x2 + 2x − 1
dx
Check!
Integrating both sides with respect to x, we have
Z
d
[(x2 + 1)y] dx =
dx
Z
x2 + 2x − 1 dx
We obtain and implicit solutition
(x2 + 1)y =
x3
+ x2 − x + C
3
To obtain an explicit solution, we divide both sides by x2 + 1:
1
y= 2
x +1
!
x3
+ x2 − x + C .
3
Solving first order linear equations
Case 3. However, the requirement a0 (x) = a01 (x) is very restrictive and in
most cases is not satisfied. To force the requirement to be satisfied,
we multiply with an appropriate function µ(x)
Definition
Solving first order linear equations
Case 3. However, the requirement a0 (x) = a01 (x) is very restrictive and in
most cases is not satisfied. To force the requirement to be satisfied,
we multiply with an appropriate function µ(x)
Definition
µ(x) is called an integrating factor.
Solving first order linear equations
Case 3. However, the requirement a0 (x) = a01 (x) is very restrictive and in
most cases is not satisfied. To force the requirement to be satisfied,
we multiply with an appropriate function µ(x)
Definition
µ(x) is called an integrating factor.
To carry the procedure, we first divide both sides by a1 (x), so that
the linear ODE becomes:
dy
a0 (x)
b(x)
+
y=
dx a1 (x)
a1 (x)
Solving first order linear equations
Case 3. However, the requirement a0 (x) = a01 (x) is very restrictive and in
most cases is not satisfied. To force the requirement to be satisfied,
we multiply with an appropriate function µ(x)
Definition
µ(x) is called an integrating factor.
To carry the procedure, we first divide both sides by a1 (x), so that
the linear ODE becomes:
dy
a0 (x)
b(x)
+
y=
dx a1 (x)
a1 (x)
Which can also be written as
dy
a0 (x)
b(x)
+ P (x)y = Q(x), where P (x) =
and Q(x) =
dx
a1 (x)
a1 (x)
Solving first order linear equations
Case 3. However, the requirement a0 (x) = a01 (x) is very restrictive and in
most cases is not satisfied. To force the requirement to be satisfied,
we multiply with an appropriate function µ(x)
Definition
µ(x) is called an integrating factor.
To carry the procedure, we first divide both sides by a1 (x), so that
the linear ODE becomes:
dy
a0 (x)
b(x)
+
y=
dx a1 (x)
a1 (x)
Which can also be written as
dy
a0 (x)
b(x)
+ P (x)y = Q(x), where P (x) =
and Q(x) =
dx
a1 (x)
a1 (x)
Definition
The equation
dy
+ P (x)y = Q(x)
dx
is called standard form of linear first order ODE (1).
Solving linear equations
Solving linear equations
Put the equation in standard form.
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
µ(x)P (x) = µ0 (x)
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
µ(x)P (x) = µ0 (x)
This is a separable equation for µ which we can solve.
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
µ(x)P (x) = µ0 (x)
This is a separable equation for µ which we can solve.
(Separate variables, integrate both sides, and solve for µ(x)):
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
µ(x)P (x) = µ0 (x)
This is a separable equation for µ which we can solve.
(Separate variables, integrate both sides, and solve for µ(x)):
µ(x) = e
R
P (x) dx
Solving linear equations
Solving linear equations
Put the equation in standard form.
Multiply both sides by µ(x):
µ(x)
dy
+ µ(x)P (x)y = µ(x)Q(x)
dx
For the condition a0 (x) = a01 (x) to be satisfied, we need
µ(x)P (x) = µ0 (x)
This is a separable equation for µ which we can solve.
(Separate variables, integrate both sides, and solve for µ(x)):
µ(x) = e
R
P (x) dx
We do not use an arbitrary constant here because we only need
ONE function µ and we may take C = 0
Solving linear equations
Solving linear equations
Multiplying both sides of the given linear ODE in standard
form by µ(x) we get
d
[µ(x)y] = µ(x)Q(x)
dx
Solving linear equations
Solving linear equations
Multiplying both sides of the given linear ODE in standard
form by µ(x) we get
d
[µ(x)y] = µ(x)Q(x)
dx
Integrating both sides with respect to x, we have
Z
µ(x)y =
µ(x)Q(x) dx + C
Solving linear equations
Solving linear equations
Multiplying both sides of the given linear ODE in standard
form by µ(x) we get
d
[µ(x)y] = µ(x)Q(x)
dx
Integrating both sides with respect to x, we have
Z
µ(x)y =
µ(x)Q(x) dx + C
Here, we add the arbitrary constant because we want ALL
solutions. (The constant must be added in this step.)
Solving linear equations
Solving linear equations
Multiplying both sides of the given linear ODE in standard
form by µ(x) we get
d
[µ(x)y] = µ(x)Q(x)
dx
Integrating both sides with respect to x, we have
Z
µ(x)y =
µ(x)Q(x) dx + C
Here, we add the arbitrary constant because we want ALL
solutions. (The constant must be added in this step.)
Dividing both sides by µ(x), we obtain an explicit solution
y(x) =
1
µ(x)
Z
µ(x)Q(x) dx + C
Example 4 – Problem #8/p.54
Example
Solve the equation
dy
y
= + 2x + 1
dx
x
Example 4 – Problem #8/p.54
Example
Solve the equation
dy
y
= + 2x + 1
dx
x
Solution
As a first step we put the equation in standard form:
dy
1
− y = 2x + 1.
dx x
Example 4 – Problem #8/p.54
Example
Solve the equation
dy
y
= + 2x + 1
dx
x
Solution
As a first step we put the equation in standard form:
dy
1
− y = 2x + 1.
dx x
Now we easily see that the equation is linear
P (x) = −
1
x
Q(x) = 2x + 1.
Example 4 – Problem #8/p.54
Example
Solve the equation
dy
y
= + 2x + 1
dx
x
Solution
As a first step we put the equation in standard form:
dy
1
− y = 2x + 1.
dx x
Now we easily see that the equation is linear
P (x) = −
1
x
We compute the integrating factor
Q(x) = 2x + 1.
Example 4 – Problem #8/p.54
Example
Solve the equation
dy
y
= + 2x + 1
dx
x
Solution
As a first step we put the equation in standard form:
dy
1
− y = 2x + 1.
dx x
Now we easily see that the equation is linear
P (x) = −
1
x
Q(x) = 2x + 1.
We compute the integrating factor
R
R
µ(x) = e P (x)dx = e
= e− ln x = eln x
−1
1
−x
dx
=
1
x
Example 4 – Problem #8/p.54
Solution
ODE in standard form :
1
dy
− y = 2x + 1.
dx x
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
1
1 dy
1
− y =2+
x dx x2
x
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
1
1 dy
1
− y =2+
x dx x2
x
d 1
1
We rewrite the left hand side as:
y =2+
dx x
x
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
1
1 dy
1
− y =2+
x dx x2
x
d 1
1
We rewrite the left hand side as:
y =2+
dx x
x
We integrate both sides
Z
Z d 1
1
y dx =
2+
dx + C
dx x
x
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
1
1 dy
1
− y =2+
x dx x2
x
d 1
1
We rewrite the left hand side as:
y =2+
dx x
x
We integrate both sides
Z
Z d 1
1
y dx =
2+
dx + C
dx x
x
We obtain:
1
y = 2x + ln |x| + C (We want ALL solutions).
x
Example 4 – Problem #8/p.54
Solution
1
dy
− y = 2x + 1.
dx x
We multiply the linear ODE in standard form by µ(x) = x1 :
ODE in standard form :
1
1 dy
1
− y =2+
x dx x2
x
d 1
1
We rewrite the left hand side as:
y =2+
dx x
x
We integrate both sides
Z
Z d 1
1
y dx =
2+
dx + C
dx x
x
1
y = 2x + ln |x| + C (We want ALL solutions).
x
We obtain an explicit solution
We obtain:
y(x) = x (2x + ln |x| + C) = 2x2 + x ln |x| + Cx
Example 5 – Problem #9/p.54
Example
Solve the equation
dr
+ r tan θ = sec θ
dθ
Example 5 – Problem #9/p.54
Example
Solve the equation
dr
+ r tan θ = sec θ
dθ
Solution
The dependent variable is r; the independent variable is θ.
Example 5 – Problem #9/p.54
Example
Solve the equation
dr
+ r tan θ = sec θ
dθ
Solution
The dependent variable is r; the independent variable is θ.
The equation is already in standard form.
Example 5 – Problem #9/p.54
Example
Solve the equation
dr
+ r tan θ = sec θ
dθ
Solution
The dependent variable is r; the independent variable is θ.
The equation is already in standard form.
P (θ) = tan θ and Q(θ) = sec θ
Example 5 – Problem #9/p.54
Example
Solve the equation
dr
+ r tan θ = sec θ
dθ
Solution
The dependent variable is r; the independent variable is θ.
The equation is already in standard form.
P (θ) = tan θ and Q(θ) = sec θ
The integrating factor is given by µ(θ) = e
R
P (θ)dθ
.
Example 5 – Problem #9/p.54
Example
dr
+ r tan θ = sec θ
dθ
Solve the equation
Solution
The dependent variable is r; the independent variable is θ.
The equation is already in standard form.
P (θ) = tan θ and Q(θ) = sec θ
The integrating factor is given by µ(θ) = e
R
µ(θ) = e
tan θdθ
R
= e− ln | cos θ| = eln |cosθ|
P (θ)dθ
−1
=
. Hence,
1
cos θ
Example 5 – Problem #9/p.54
Example
dr
+ r tan θ = sec θ
dθ
Solve the equation
Solution
The dependent variable is r; the independent variable is θ.
The equation is already in standard form.
P (θ) = tan θ and Q(θ) = sec θ
The integrating factor is given by µ(θ) = e
R
µ(θ) = e
tan θdθ
R
= e− ln | cos θ| = eln |cosθ|
P (θ)dθ
−1
=
. Hence,
1
cos θ
We have dropped the absolute value because we are looking
for solution where cos θ > 0. cos θ cannot be zero or less than
zero because then µ will not exist.
Example 5 – Problem #9/p.54
Solution
We multiply both sides by µ(θ) =
1
cos θ :
1 dr
1
+r
tan θ = sec2 θ
cos θ dθ
cos θ
Example 5 – Problem #9/p.54
Solution
We multiply both sides by µ(θ) =
1
cos θ :
1 dr
1
+r
tan θ = sec2 θ
cos θ dθ
cos θ
Hence, the left-hand side can be written as
d
1
r = sec2 θ
dθ cos θ
Example 5 – Problem #9/p.54
Solution
We multiply both sides by µ(θ) =
1
cos θ :
1 dr
1
+r
tan θ = sec2 θ
cos θ dθ
cos θ
Hence, the left-hand side can be written as
d
1
r = sec2 θ
dθ cos θ
We integrate both sides, recalling that sec θ = 1/ cos θ
Z Z
1
r dθ = sec2 θdθ
cos θ
Example 5 – Problem #9/p.54
Solution
We multiply both sides by µ(θ) =
1
cos θ :
1 dr
1
+r
tan θ = sec2 θ
cos θ dθ
cos θ
Hence, the left-hand side can be written as
d
1
r = sec2 θ
dθ cos θ
We integrate both sides, recalling that sec θ = 1/ cos θ
Z Z
1
r dθ = sec2 θdθ
cos θ
Using the Table of Integrals in the back of the book, we get
1
r = tan θ + C
cos θ
Example 5 – Problem #9/p.54
Solution
We multiply both sides by µ(θ) =
1
cos θ :
1 dr
1
+r
tan θ = sec2 θ
cos θ dθ
cos θ
Hence, the left-hand side can be written as
d
1
r = sec2 θ
dθ cos θ
We integrate both sides, recalling that sec θ = 1/ cos θ
Z Z
1
r dθ = sec2 θdθ
cos θ
Using the Table of Integrals in the back of the book, we get
1
r = tan θ + C
cos θ
The explicit solution then is:
r(θ) = sin θ + C cos θ
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
dy
3
sin x
+ y = 2 − 3x
dx x
x
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
dy
3
sin x
+ y = 2 − 3x
dx x
x
Thus, we have
P (x) =
3
x
Q(θ) =
sin x
− 3x
x2
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
dy
3
sin x
+ y = 2 − 3x
dx x
x
Thus, we have
The integrating factor
P (x) =
3
x
µ(x) = e
Q(θ) =
R
P (x)dx
sin x
− 3x
x2
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
dy
3
sin x
+ y = 2 − 3x
dx x
x
Thus, we have
P (x) =
The integrating factor
Hence,
R
µ(x) = e
3
x
µ(x) = e
3
dx
x
Q(θ) =
R
sin x
− 3x
x2
P (x)dx
3
= e3 ln x = eln x = x3
Example 6 – Problem #14/p.54
Example
Solve the equation
x
sin x
dy
+ 3(y + x2 ) =
dx
x
Solution
We first express the equation in standard form (divide both
sides by x and then subtract 3x from both sides).
dy
3
sin x
+ y = 2 − 3x
dx x
x
Thus, we have
P (x) =
The integrating factor
Hence,
R
µ(x) = e
3
x
µ(x) = e
3
dx
x
Q(θ) =
R
sin x
− 3x
x2
P (x)dx
3
= e3 ln x = eln x = x3
We have dropped the absolute value because we are looking
for solution where x > 0.
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
x3
dy
+ 3x2 y = x sin x − 3x4
dx
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
x3
dy
+ 3x2 y = x sin x − 3x4
dx
Hence, the left-hand side can be written as
d 3 x y = x sin x − 3x4
dx
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
x3
dy
+ 3x2 y = x sin x − 3x4
dx
Hence, the left-hand side can be written as
d 3 x y = x sin x − 3x4
dx
Z
Z
d
We integrate both sides:
x3 y dx = (x sin x − 3x4 )dx
dx
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
x3
dy
+ 3x2 y = x sin x − 3x4
dx
Hence, the left-hand side can be written as
d 3 x y = x sin x − 3x4
dx
Z
Z
d
We integrate both sides:
x3 y dx = (x sin x − 3x4 )dx
dx
R
Using the Table of Integrals in the back of the book for x sin xdx,
we get
3
x3 y = sin x − x cos x − x5 + C
5
Example 6 – Problem #14/p.54
Solution
We multiply both sides by µ(x) = x3 :
x3
dy
+ 3x2 y = x sin x − 3x4
dx
Hence, the left-hand side can be written as
d 3 x y = x sin x − 3x4
dx
Z
Z
d
We integrate both sides:
x3 y dx = (x sin x − 3x4 )dx
dx
R
Using the Table of Integrals in the back of the book for x sin xdx,
we get
3
x3 y = sin x − x cos x − x5 + C
5
3
We divide both sides by x to obtain the explicit solution:
y(x) =
sin x − x cos x 3 2
C
− x + 3
3
x
5
x
Initial Value Problems
When we solve IVPs of the form
(
y 0 = f (x, y)
y(x0 ) = y0
Initial Value Problems
When we solve IVPs of the form
(
y 0 = f (x, y)
y(x0 ) = y0
We first solve the ODE and obtain a one-parameter family of
solutions (or general solution).
Initial Value Problems
When we solve IVPs of the form
(
y 0 = f (x, y)
y(x0 ) = y0
We first solve the ODE and obtain a one-parameter family of
solutions (or general solution).
We use the initial condition to determine the unknown
constant.
Initial Value Problems
When we solve IVPs of the form
(
y 0 = f (x, y)
y(x0 ) = y0
We first solve the ODE and obtain a one-parameter family of
solutions (or general solution).
We use the initial condition to determine the unknown
constant.
It is easier to do that if we use the initial condition to
determine the unknown constant with the implicit form of the
solution as obtain from the method of separation of variables.
Initial Value Problems
When we solve IVPs of the form
(
y 0 = f (x, y)
y(x0 ) = y0
We first solve the ODE and obtain a one-parameter family of
solutions (or general solution).
We use the initial condition to determine the unknown
constant.
It is easier to do that if we use the initial condition to
determine the unknown constant with the implicit form of the
solution as obtain from the method of separation of variables.
We demonstrate this in the following examples.
Example 7 – Problem#17/p.54
Example
Solve the following initial value problem (IVP)
( dy
− x1 y = xex
y(1) = e − 1
dx
Example 7 – Problem#17/p.54
Example
Solve the following initial value problem (IVP)
( dy
− x1 y = xex
y(1) = e − 1
dx
Solution
The equation is already in standard form. We have
P (x) = −
1
x
Q(x) = xex
Example 7 – Problem#17/p.54
Example
Solve the following initial value problem (IVP)
( dy
− x1 y = xex
y(1) = e − 1
dx
Solution
The equation is already in standard form. We have
P (x) = −
1
x
The integrating factor µ(x) is:
Q(x) = xex
µ(x) = e−
R
P (x)dx
Example 7 – Problem#17/p.54
Example
Solve the following initial value problem (IVP)
( dy
− x1 y = xex
y(1) = e − 1
dx
Solution
The equation is already in standard form. We have
P (x) = −
1
x
The integrating factor µ(x) is:
µ(x) = e−
R
1
x
dx
Q(x) = xex
µ(x) = e−
R
−1
= e− ln x = eln x
P (x)dx
=
1
x
Example 7 – Problem#17/p.54
Solution
We multiply both sides of the linear equation by µ(x) = x1 :
Example 7 – Problem#17/p.54
Solution
We multiply both sides of the linear equation by µ(x) = x1 :
1 dy
y
−
= ex
x dx x2
Example 7 – Problem#17/p.54
Solution
We multiply both sides of the linear equation by µ(x) = x1 :
1 dy
y
−
= ex
x dx x2
We rewrite the left-hand side as a derivative of a product:
d
dx
1
y = ex
x
Example 7 – Problem#17/p.54
Solution
We multiply both sides of the linear equation by µ(x) = x1 :
1 dy
y
−
= ex
x dx x2
We rewrite the left-hand side as a derivative of a product:
d
dx
1
y = ex
x
We integrate both sides.
Z
d
dx
1
y dx =
x
Z
ex dx
Example 7 – Problem#17/p.54
Solution
We multiply both sides of the linear equation by µ(x) = x1 :
1 dy
y
−
= ex
x dx x2
We rewrite the left-hand side as a derivative of a product:
d
dx
1
y = ex
x
We integrate both sides.
Z
d
dx
1
y dx =
x
Z
ex dx
In the left-hand side, we have derivative after integral
1
y = ex + C
x
Example 7 – Problem#17/p.46
Solution
Next we use the initial condition to evaluate C. We have
x = 1, y = e − 1.
Example 7 – Problem#17/p.46
Solution
Next we use the initial condition to evaluate C. We have
x = 1, y = e − 1.
Substituting in the implicit solution we get
e−1=e+C
Hence,
−1 = C
Example 7 – Problem#17/p.46
Solution
Next we use the initial condition to evaluate C. We have
x = 1, y = e − 1.
Substituting in the implicit solution we get
e−1=e+C
Hence,
−1 = C
Finally, the unique solution is
1
y = ex − 1
x
Example 7 – Problem#17/p.46
Solution
Next we use the initial condition to evaluate C. We have
x = 1, y = e − 1.
Substituting in the implicit solution we get
e−1=e+C
Hence,
−1 = C
Finally, the unique solution is
1
y = ex − 1
x
We may also obtain the explicit solution:
y = x(ex − 1)
Example 8 – Problem#22/p.54
Example
Solve the following initial value problem (IVP)
(
dy
+ cos xy = x sin x
sin x dx
π
y( 2 ) = 2
Example 8 – Problem#22/p.54
Example
Solve the following initial value problem (IVP)
(
dy
+ cos xy = x sin x
sin x dx
π
y( 2 ) = 2
Solution
The ODE is NOT in standard form BUT a0 (x) = a01 (x).
Example 8 – Problem#22/p.54
Example
Solve the following initial value problem (IVP)
(
dy
+ cos xy = x sin x
sin x dx
π
y( 2 ) = 2
Solution
The ODE is NOT in standard form BUT a0 (x) = a01 (x).
We can directly rewrite the left-hand side as a derivative of a
product:
d
[(sin x)y] = x sin x
dx
Example 8 – Problem#22/p.54
Example
Solve the following initial value problem (IVP)
(
dy
+ cos xy = x sin x
sin x dx
π
y( 2 ) = 2
Solution
The ODE is NOT in standard form BUT a0 (x) = a01 (x).
We can directly rewrite the left-hand side as a derivative of a
product:
d
[(sin x)y] = x sin x
dx
We integrate both sides.
Z
d
((sin x)y) dx =
dx
Z
x sin xdx
Example 8 – Problem#22/p.46
Solution
In the left-hand side, we have derivative after integral. For the
right-hand side we use the Table of Integrals to integrate
((sin x)y) = sin x − x cos x + C
Example 8 – Problem#22/p.46
Solution
In the left-hand side, we have derivative after integral. For the
right-hand side we use the Table of Integrals to integrate
((sin x)y) = sin x − x cos x + C
Next we evaluate C. From the IC, we have x = π2 , y = 2.
Example 8 – Problem#22/p.46
Solution
In the left-hand side, we have derivative after integral. For the
right-hand side we use the Table of Integrals to integrate
((sin x)y) = sin x − x cos x + C
Next we evaluate C. From the IC, we have x = π2 , y = 2.
Substituting in the implicit solution we get
2 = 1 + C =⇒ 1 = C
Example 8 – Problem#22/p.46
Solution
In the left-hand side, we have derivative after integral. For the
right-hand side we use the Table of Integrals to integrate
((sin x)y) = sin x − x cos x + C
Next we evaluate C. From the IC, we have x = π2 , y = 2.
Substituting in the implicit solution we get
2 = 1 + C =⇒ 1 = C
Finally, the unique solution is
(sin x)y = sin x − x cos x + 1
Example 8 – Problem#22/p.46
Solution
In the left-hand side, we have derivative after integral. For the
right-hand side we use the Table of Integrals to integrate
((sin x)y) = sin x − x cos x + C
Next we evaluate C. From the IC, we have x = π2 , y = 2.
Substituting in the implicit solution we get
2 = 1 + C =⇒ 1 = C
Finally, the unique solution is
(sin x)y = sin x − x cos x + 1
We may also obtain the explicit solution:
y(x) =
sin x − x cos x + 1
sin x