Chapter 1 First-Order Differential Equations

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Chapter 1 First-Order Differential Equations

• Shurong Sun
• University of Jinan
• Semester 1, 2010-2011

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Analogous to a course in algebra and
trigonometry, where a good amount of time is
spent solving equations such as
for the unknown variable x.
In this course one of our tasks will be to solve
differential equations such as
for the unknown function yy(t).
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1.1 Separable Differential Equations

• Many first order differential equations can be
  written as

• We can solve these equations by integrating

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Separable Differential Equation
A first order differential equation of the form
is called a separable differential equation.
I.
Set
and multiply by
to obtain
Integrating, we have
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II.
then
is one solution to the equation
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Separable Equation Example
Example Solve the equation
Solution
This is a separable equation.
Separate variables and rewrite the equation in
the form
or
Integrating, we have
Solving for y , we obtain the solution in
explicit form as
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Separable Equation Example

• Solve the differential equation

Solution
We first rewrite the DE in fractional form
Separate variables and rewrite the equation in
the form
Integrating, we get
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Example Solve the IVP
Solution
Separate variables and rewrite the equation
in the form
or
Integrating, we have
Solving for y , we obtain the solution in
explicit form as
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Note that y1 is also a solution to the equation.
So the general solution to the equation is
is an arbitrary constant.
Applying the initial condition directly, we have
or
Thus
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Example Solve the equation
Solution
Separate variables and rewrite the equation
in the form
Integrating, we have
or
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Example- Newton s Law of Cooling
A copper sphere is heated to
At time t0 it is
then placed in water that is maintained at
After 3 minutes the sphere s temperature is
Derive an equation for the temperature T of the
ball
as a function of time t.

• Newton s law of cooling implies

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Example- Newton s Law of Cooling

• Step 1 - Separate variables

• Step 2 - Integrate

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Example- Newton s Law of Cooling

• Step 3 - Determine a particular solution

• Step 4 - Determine k

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Example
A hemispherical bowl has top radius 4 ft and at
time t0 is full of water. At that moment a
circular hole with diameter 1 in. is opened in
the bottom of the tank. How long will it take for
all the water to drain from the tank?
Solution
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1.2 Reduction to Separable Form

• Sometimes first order differential equations can
  be made separable by a simple change of variable.
  Consider equations of the form

(1) Homogeneous Equation
Remark
is homogeneous iff
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• This suggests we set

• Writing the differential equation in terms of u
  gives

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• Separating variables gives

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Example Solve
(6)
Solution We express (6) in the derivative form
then we see that the equation (6) is homogeneous.
Now let
and recall that
With these substitution, equation (6) becomes
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The above equation is separable, and , on
separating the
variables and integrating, we obtain
Hence,
Also note that x0 is a solution.
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(2) Equations of the Form
Set
Then the equation is transformed
into a separable one.
Example Solve
(8)
Solution
Set
Then
and so
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Substituting into (8) yields
Solving this separable equation, we obtain
from which it follows that
Finally, replacing z by x-y yields
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(3)
where the
are constants.
I.
In this case, the equations can be reduced to
the form
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II.
Then the system of equations
has a unique solution
The above DE can be written in the form
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which yields the DE
Homogenous Equation
after the translation of axes of the form
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Example Solve
Solution
From
we obtain
Hence, we let
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The differential equation for v is
or
The above equation is homogenous, so we let
Then
or
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Separating variables gives
from which it follows that
When we substitute back in for z, u, and v, we
find
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1.2 Linear First-Order DE
(1) Definition Linear Equation
A first-order differential equation of the form
(1)
is said to be linear equation.
When
the linear equation is said
to be homogeneous
otherwise, it is non-homogenous
or inhomogenous.
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(2) Standard Form
By dividing both sides of (1) by the lead
coefficient
we obtain a more useful form, the stand
form, of a linear equation
(2)
We seek a solution of (2) on an interval I for
which
both function P and f are continuous.
(3) Variation of parameters
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I. The Homogeneous Equation
(3)
This is a separable equation.
Writing (3) as
and integrating .
Solving for y gives
II. The Nonhomogeneous Equation
(2)
Method of variation of constants
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II. The Nonhomogeneous Equation
(2)
Method of variation of constants
The basic idea is the constant C in the general
solution of the homogeneous equation (3) is
replaced by a function C(x). The calculation of
an appropriate choice of C(x) gives a solution of
the nonhomogeneous equation (2) .
Substituting
into (2) gives
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so
Hence
and
(4)
Thus if (2) has a solution, it must be of form
(4).
Conversely, it is easy to verify that (4)
constitutes one-parameter family of solutions of
equation (2),
that is a general solution of equation (2).
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Example
Finding the general solution to
Solution
We write the differential equation in standard
form
The associated homogeneous equation is
Separating variable, we find
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So the general solution to the homogeneous
equation is
substituting
By the method of variation of parameter,
into the equation gives
So
Thus the general solution to
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is
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Method of Integrating factor
An integrating factor
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Example
Find the general solution to
Solution We write the differential equation in
the
standard form
Here
So the integrating factor is
Multiplying the original differential equation by
we get
or
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We now integrate both sides and solve for y to
find
or
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Find the general solution to
Solution
Here
So
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Solve
Solution Here
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Bernoulli Equations
(9)
Remark when n0,1, equation (9) is also a
linear equation.
For
dividing equation (9) by
yields
(10)
Taking
we find via the chain rule that
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and so equation (10) becomes
Linear equation
Example Solve
(11)
Solution This is a Bernoulli equation with n3,
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We make the substitution
Since
the transformed equation is
This is a linear equation, its solution is
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Substituting
give the solution
Not included in the last equation is the solution
that was lost in the process of dividing
(11) by
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1.3 Exact Differential Equations and Integrating
factors
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Exact Differential Equations

• Recall that the total or exact differential of a
  function u(x,y) is

• We will use the concept of exactness to study
  differential equations of the form

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• Compare the following

We say an ODE is exact if there exists a function
such that
That is
Note this means that we can now write our ODE as
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Note this means that we can now write our ODE as
In this case, its solution (in implicit form) is
given by
Remember we need
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Testing for Exactness

• In practice we often don t know about u, only
  about M and N and it is hard to check that

• We want a technique of testing for exactness
  based on knowing M and N, not u

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• Let s check the derivatives of M and N

If M(x,y) and N(x,y) are continuous functions and
have continuous first partial derivatives on some
simply connected region of xy-plane, then

• A necessary and sufficient condition for
  exactness is

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Solving Exact Equations

• Once we have checked the equation is exact it can
  be readily solved by evaluating either

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Exact Equations Example
Example Solve
Solution
the equation given is exact.
From
we have that
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Next we take the partial derivative of u with
respect
to y and substitute
for N
Here we drop the constant of integration that
technically should be present in g(y) since it
will just get absorbed into the constant we pick
up in the next step.
Hence
So, the implicit solution to the differential
equation is
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Exact Equations Example
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Exact Equations Example

• Solve for k(x) and then u

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Example Find the solution for the following IVP.
Solution
Let s identify M and N and check that it s
exact.
So, it s exact.
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With the proper simplification integrating the
second one isn t too bad. However, the first is
already set up for easy integration so let s do
that one.
Differentiate with respect to y and compare to N.
So, we get
Recall that actually h(y) k, but we drop the k
because it will get absorbed in the next step.
That gives us h(y) 0.
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Therefore, we get
The implicit solution is then
Applying the initial condition gives 1 C
The implicit solution is then
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Remember Exact Equations

• We can check whether an equation in this form is
  exact by checking if

(Necessary and sufficient)

• Alternate method we can often rearrange linear
  first order ODEs into the form

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Solve the following DE
Solution
Here
We first check to see if we have an exact
equation.
Since
it s exact.
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Note that
So the general solution is
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Example. Find the general solution to
Step 1 Check to see
So, it is exact.
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Then,
So, we get
as the general solution.
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Example. Solve the homogeneous DE
Solution This equation can be written in the form
which is an exact equation.
In this case, the solution in implicit form is
i.e. ,
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Integrating Factors

• If the equation is not exact we can consider

multiplying it by a function
such that the new equation is exact
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Finding Integrating Factors

• For exactness we require

This equation can be simplified in special cases,
two.
of which we treat next
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• If we choose

where
is just a function of x.
Let
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Now solve for
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Conversely, if
is just a function of x.
Let
Then
is an integrating factor for Equation.
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In a similar fashion, if equation has an
integrating
factor that depends only y, then
is just a function of y.
Conversely, if
is just a function of y.
Let
Then
is an integrating factor for Equation.
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Integrating Factors Example
Example Solve
Solution
The equation is not exact.
We compute
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So an integrating factor for the equation is
given by
When we multiplying by
we get the exact equation
Solving this equation, we obtain the implicit
solution
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Note that the solution x0 is lost in multiplying
by
Hence,
and
are solutions to the given equation.
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Linear Differential Equations

• Recall a differential equation is linear if it
  can be written
• If q(x)0 the equation is homogeneous, otherwise
  the equation is nonhomogeneous.

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Homogeneous Linear Case

• Separating variables and integrating

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Nonhomogeneous Linear Case

• Nonhomogeneous linear equations can be solved
  using the integrating factor method. First
  rewrite the differential equation as

i.e.
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Finding the integrating factor

• To find the integrating factor we first compute

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Nonhomogeneous Linear Case

• We can now find an integrating factor

• Now multiply the differential equation by the
  integrating factor

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Nonhomogeneous Linear Case
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Integrating Factors Example
Example Solve
Solution
Here
The equation is not exact.
We compute
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So an integrating factor for is given by
When we multiplying by
we get the exact equation
Solving this equation, we obtain
as the solution in implicit form.
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In general, integrating factors are
difficult to uncover. If a differential
equation does not have one of the forms given
above, then a search for an integrating factor
likely will not be successful, and other methods
of solution are recommended.
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Example Solve
Solution Rewriting this
equation in differential form,
we have
This differential equation is not exact, and
no integrating factor is immediately apparent.
Note however, that if terms are
strategically regrouped, the DE can be rewritten
as
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(1)
The first group of terms has many integrating
factors (see Table 2). One of these factors,
namely
is an integrating factor for the entire equation.
Multiplying (1) by
we find
(2)
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Since (2) is exact, it can be solved using the
steps described previously.
Alternatively, we note from Table 1,
so that (2) can be rewrite as
Integrating both sides of this last equation, we
find
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or equivalently,
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Solve
Solution Here
and, since
The differential equation is not exact.
Since
is a function of y alone.
we have an integrating factor
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Multiplying the given DE by
we obtain the exact equation
or equivalently,
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Integrating both sides of this last equation, we
find
(ii) Note that the differential equation can be
rewritten as
The first group of terms has many integrating
factors (see Table 1). One of these factors,
namely
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is an integrating factor for the entire equation.
Multiplying (1) by
we find
Integrating both sides of this last equation, we
find
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(iii) Rewriting this equation in the derivative
form,
(3)
then we see that the equation (3) is homogeneous.
Now let
and recall that
With these substitution, equation (3) becomes
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The above equation is separable, and , by
separating the
variables and integrating, we obtain
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(iv) We rewrite the differential equation in the
form
which is linear equation.
Its solution is
Also note that x0 is a solution.
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Table 1
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Table 2
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Exercise Solve
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