First Order Differential Equations
Second Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
First Order Differential Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
Example 28
Solve the following IVP .
2xy 2 + 4 = 2(3 − x 2 y )y 0 ;
y (1) = 8
Solution
Here, we first need to put the differential equation into proper
form before proceeding. Recall that it needs to be
2xy 2 + 4 − 2(3 − x 2 y )y 0 = 0
2xy 2 + 4 + 2(x 2 y − 3)y 0 = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
So we have the following
M = 2xy 2 + 4
N = 2x 2 y − 6
My = 4xy = Nx
So, the differential equation is exact according to the test. Now,
how do we actually find ψ(x, y ) ? Well recall that
ψ(x, y )x = M
ψ(x, y )y = N
We can use either of these to get a start on finding ψ(x, y ) by
integrating as follows. We will integrate N with repect to y and we
will ad a ”constant” of integration .
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
Z
ψ(x, y ) =
Ndy + g (x)
So, I will use the second one.
Z
ψ(x, y ) = (2x 2 y − 6)dy = x 2 y 2 − 6y + g (x)
This time, as opposed to the previous example, our constant of
integration must be a function of x since we integrated with
respect to y . Now differentiate with respect to x and compare this
to M.
ψ(x, y )x = 2xy 2 + g 0 (x) = M = 2xy 2 + 4
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
From this we can see that
h0 (y ) = 4 =⇒ g (x) = 4x
k =?
Writing everything down gives us the following for ψ(x, y )
ψ(x, y ) = x 2 y 2 − 6y + 4x
the implicit solution is then
ψ(x, y ) = c =⇒ x 2 y 2 − 6y + 4x = c
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
Applying the initial conditions
64 − 48 − 4 = c =⇒ c = 12
The complete (including the constant)implicit solution is then.
x 2 y 2 − 6y + 4x = 12
Now, this is quadratic in y and so we can solve for y (x) by using
the quadratic formula.
p
6 ± 36 − 4(x 2 )(4x − 6)
y=
2x 2
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
√
36 + 48x 2 − 16x 3
2x 2
√
3 ± 9 + 12x 2 − 4x 3
y=
x2
Reapplying the initial condition shows that this time we need the
plus sign. Therefore, the explicit solution
√
3 + 9 + 12x 2 − 4x 3
y=
x2
y=
6±
Dr. Marco A Roque Sol
Ordinary Differential Equations
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Integrating Factor
A function µ(x, y ) is an integrating factor of the equation
M(x, y ) + N(x, y )y 0 (x) = 0
if the ODE
µ(x, y )M(x, y ) + µ(x, y )N(x, y )y 0 (x) = 0
is exact, that means
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
∂µ(x, y )M(x, y )
∂µ(x, y )N(x, y )
=
∂y
∂x
The above condition is a partial differential equation, that in
general is quite dificult to solve. However, if we assume that
µ(x, y ) = µ(x) then the above equation becomes
∂µ(x)N(x, y )
∂M(x, y )
=
∂y
∂x
My − Nx
dµ(x)
=
µ(x)
dx
N
µ(x)
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
My − Nx
depends only on x , then we have a linear
N
differential equation por µ
Now if
dµ(x)
+ p(x)µ(x) = 0
dx
with p(x) given by
p(x) =
Dr. Marco A Roque Sol
My − Nx
N
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
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Example 29
Find the solution of the ODE
(3x 2 y + 2xy + y 3 )dx + (x 2 + y 2 )dy = 0
Here
M = 3x 2 y + 2xy + y 3
N = x2 + y2
Since
My = 3x 2 + 2x + 3y 2 6= 2x = Nx
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
Therefore, this equation is not exact. We need to find an
integrating factor , but the combination
M y − Nx
3x 2 + 2x + 3y 2 6= 2x
3(x 2 + y 2 )
=
=
=3
N
x2 + y2
x2 + y2
is independent of y , thus µ satisfies the differential equation
µ0 (x) + 3µ(x) = 0
whose solution is
µ=e
R
3dx
Dr. Marco A Roque Sol
(x) = e 3x
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
We can now employ this µ(x) as an integrating factor to construct
a general solution of
e 3x (3x 2 y + 2xy + y 3 )dx + e 3x (x 2 + y 2 )dy = 0
which, by construction, must be exact. So we seek a function ψ
such that
ψ(x, y )x = e 3x (3x 2 y + 2xy + y 3 )
Dr. Marco A Roque Sol
and
ψ(x, y )y = e 3x (x 2 + y 2 )
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Exact Equations and Integrating Factors
Exact Equations and Integrating Factors
Integrating the second of the above equations with respect to y
1
ψ(x, y ) = x 2 ye 3x + y 3 e 3x + g (x)
3
but remembering that
ψ(x, y )x = e 3x (3x 2 y + 2xy + y 3 )
we find that g (x) = C , and the implicit form of the solution is
1
x 2 ye 3x + y 3 e 3x = c
3
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Homogeneus Equations
Solutions of Linear Homogeneus Systems
Second Order Differential Equations
In this chapter we will be looking exclusively at linear second order
differential equations. The most general linear second order
differential equation is in the form.
p(t)y 00 + q(t)y 0 + r (t)y = g (t)
In fact, we will rarely look at non-constant coefficient linear second
order differential equations. In the case where we assume constant
coefficients we will use the following differential equation.
ay 00 + by 0 + cy = g (t)
Where possible we will use the first equation just to make the
point that certain facts, theorems, properties, and/or techniques
can be used with the non-constant form.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Homogeneus Equations
Solutions of Linear Homogeneus Systems
Homogeneus Equations
However, most of the time we will be using the second one as it
can be fairly difficult to solve second order non-constant coefficient
differential equations.
Initially we will make our life easier by looking at differential
equations with g (t) = 0. When g (t) = 0 we call the differential
equation homogeneous and when g (t) 6= 0 we call the differential
equation nonhomogeneous.
So, lets start thinking about how to go about solving a constant
coefficient, homogeneous, linear, second order differential equation.
It is probably best to start off with an example. This example will
lead us to a very important fact that we will use in every problem
from this point on. The example will also give us clues into how to
go about solving these in general.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Homogeneus Equations
Solutions of Linear Homogeneus Systems
Homogeneus Equations
Example 30
Determine some solutions to
y 00 − 9y = 0
Solution
We can get some solutions here simply by inspection. We need
functions whose second derivative is 9 times the original function.
So, it looks like the following two functions are solutions.
y (t) = e 3t
and
y (t) = e −3t
We’ll leave it to you to verify that these are in fact solutions.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Homogeneus Equations
Solutions of Linear Homogeneus Systems
Homogeneus Equations
These two functions are not the only solutions to the differential
equation however. Any of the following are also solutions to the
differential equation.
y (t) = −9e 3t
and
y (t) = 123e 3t
y (t) = 56e −3t
and
y (t) =
y (t) = 7e 3t − 6e −3t
and
14 −3t
e
9
y (t) = −92e 3t − 16e −3t
In fact if you think about it any function that is in the form
y (t) = c1 e 3t + c2 e −3t
will be a solution to the differential equation.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Second Order Differential Equations
Homogeneus Equations
Solutions of Linear Homogeneus Systems
Solutions of Linear Homogeneus Systems
The above example leads us to a very important fact that we will
use in practically every problem in this chapter.
Principle of Superposition
If y1 and y2 are two solutions to a linear homogeneous differential
equation then so is
y (t) = c1 y1 + c2 y2
Note that we didn’t include the restriction of constant coefficient
or second order in this. This will work for any linear homogeneous
differential equation.
If we further assume second order and one other condition, we can
go a step further.
Dr. Marco A Roque Sol
Ordinary Differential Equations