Differential Equations
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Transcript Differential Equations
First, a little review:
Consider:
then:
y x2 3
y x2 5
or
y 2 x
y 2 x
It doesn’t matter whether the constant was 3 or -5, since
when we take the derivative the constant disappears.
However, when we try to reverse the operation:
Given: y 2 x
y x2 C
find
y
We don’t know what the
constant is, so we put “C” in
the answer to remind us that
there might have been a
constant.
If we have some more information we can find C.
Given: y 2 x and y 4 when x 1 , find the equation for y .
y x2 C
4 1 C
2
3C
This is called an initial value
problem. We need the initial
values to find the constant.
y x2 3
An equation containing a derivative is called a differential
equation. It becomes an initial value problem when you
are given the initial condition and asked to find the original
equation.
Integrals such as
4
1
x2 dx are called definite integrals
because we can find a definite value for the answer.
4
1
2
x dx
4
1 3
x C
3
1
1 3
1 3
4 C 1 C
3
3
Recall how the constant
always cancels when finding
a definite integral, so we
leave it out!
64
1
63
21
C C
3
3
3
Integrals such as
x dx
2
are called indefinite integrals
because we can not find a definite value for the answer.
2
x
dx
1 3
x C
3
When finding indefinite
integrals, we always
include the “plus C”.
Many of the integral formulas are listed on page 307. The
first ones that we will be using are just the derivative
formulas in reverse.
Formulas like 2, 3, and 4 might not make sense at first but
you will still need them for the homework assignment.
How they work will be explained in section 6.2.
On page 308, the book shows a technique to graph the
integral of a function using the numerical integration
function of the calculator (fnInt).
x
y t sin t dt
or
0
This is extremely slow and usually not worth the trouble.
Many of the integral formulas are listed on page 307. The
first ones that we will be using are just the derivative
formulas in reverse.
Formulas like 2, 3, and 4 might not make sense at first but
you will still need them for the homework assignment.
How they work will be explained in section 6.2.
With these formulas, you can now handle the problems in
Assignment 6.1…
With these formulas, you can now handle the problems in
Assignment 6.1…
Some differential equations can be expressed as the
product of a function of x and a function of y.
dy
g x h y
dx
How would we solve
for y if they are are part
of the same derivative?
dy
2 xy 2
dx
dy
2 x dx
2
y
h y 0
for reasons you’ll see
in a minute.
Multiply both sides by dx and divide
both sides by y2 to separate the
variables. (Assume for now that y2 is
never zero.)
y 2 dy 2 x dx
Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y
dx
h y 0
Example:
dy
2 xy 2
dx
dy
2 x dx
2
y
y 2 dy 2 x dx
Multiply both sides by dx and divide
both sides by y2 to separate the
variables. (Assume for now that y2 is
never zero.)
Separation of Variables comes down to
this: Get the y’s with the dy’s and the
x’s with the dx’s.
Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y
dx
dy
2 xy 2
dx
dy
2 x dx
2
y
y 2 dy 2 x dx
h y 0
2
y
dy 2 x dx
1
y C1 x C2
2
1
x2 C
y
1
2
y
x C
Combined
constants of
integration
1
y 2
x C
Separable Differential Equations
This is called a general solution of the differential
equation. If you were given an initial value and asked
you to solve for y, then you’re being asked to find a
particular solution.
1
y 2
x C
For example, if (0, 1) were an initial value, find the particular solution.
Answer:
1
y 2
x 1
particular solution
Separable Differential Equations
This is called a general solution of the differential
equation. If you were given an initial value and asked
you to solve for y, then you’re being asked to find a
particular solution.
1
y 2
x 1
The domain of this solution would be…
…only an open interval that contains the initial point (0,1)
But why?
This is where a slope field graph can help you
understand this.
1
Below is the graph of y 2
x 1
Since we were given the
initial value (0, 1), the only
solution we know for sure
exists is on an interval that
contains that point.
y
You can think of this in
the same way you
learned what made a
function continuous:
When you draw
the solution on a
slope field, you
should never have
to pick your pencil
up off the graph.
So the graph should
look like…
x
1
y 2
x 1
…this
−1 < x < 1
y
…because this is
the interval that
contains the point
(0, 1)
And since this
function is undefined
at x = ±1
x
Separable Differential Equations
This is called a general solution of the differential
equation. If you were given an initial value and asked
you to solve for y, then you’re being asked to find a
particular solution.
1
y 2
x 1
So the domain of this solution would be…
−1 < x < 1
Remember this
differential equation?
Sketch an
approximate curve
for y given the
initial value (0,1).
•
Now sketch an
approximate
curve
for y given the initial
value (–1, –1).
y' 2 xy
x y y
0
0
1
1
0
1
1
2
0
0
2
4
2 0
2 1
0
4
2 2
8
-1 0
0
-1 1 -2
Find a general solution for y.
dy
2 x dx
y
dy
2 xy
dx
y' 2 xy
Remember this
differential equation?
dy
2 x dx
y
ln y x c
2
y e
x2 c
y e
x c
2
y e e
c x2
We can do better than ±ec
Since e is a constant and c is a constant…
e A
c
Find a general solution for y.
dy
2 x dx
y
dy
2 xy
dx
y' 2 xy
Remember this
differential equation?
dy
2 x dx
y
ln y x c
2
y e e
c x2
Since
ec A
y Ae
x2
Remember this
differential equation?
Solve for y given
the initial value
(0,1).
y' 2 xy
ye
x2
We can do better than ±ec
e A
c
Since e is a constant and c is a constant…
We’ve seen
how it was
done with A
Now let’s compare methods: A to ±ec
Solve for y given
the initial value
(0,1).
1 e
c
y e e
c x2
ln 1 c
ye
x2
c 02
1 e e
ye e
ln1 x 2
Which is the answer we
got before
Remember this
differential equation?
Solve for y given
the initial value
(0,1).
Now solve for y
given the initial
value
(–1,
–1).
y' 2 xy
ye
y e
x2
x 2 1
What would be
an approximate
sketch of these
two graphs?
On your calculator,
graph the slope
field and both
solution curves
that you found.
p