Download Differential Equation

Ch. 7 – Differential Equations
and Mathematical Modeling
7.1 – Slope Fields and Euler’s
Method
• Differential Equation: An equation involving a derivative
dy
• Ex: Find the particular solution to dx  2 x  3 x  1
graph passes through the point (4, 2).
whose
– First, find the indefinite integral of the derivative above (it will include a
C): 2 3x3/2
yx 
3
 xC
 x 2  2 x3/2  x  C
2
– Next,
solve
2
3/2 for C using the point provided:
2  4  2(4)
4C
– Final2 answer:
3/2
y  x  2x
 x2
2  16 16  4  C
2  C
Solving Initial Value Problems!
• Rewrite the differential equation f’(x) as
an FTC problem, incorporating the
initial value
Ind. Variable
(usually x)
x
f ( x)  3   f '(t ) dt 
7
Y-value of
initial value
x
3   e dt
t 2
7
Variable of Integration
(not ind. Variable)
X-value of
initial value
Differential Equation
(switch variable)
Story Time!
• Ex: A rocket is launched with an initial velocity of 160 m/s from
a platform 3 m above the ground. Assuming the acceleration
due to gravity is a constant -9.8 m/s2, …
– Find the velocity as a function of time (t).
• Make an equation for a(t), then find the integral of a(t)!
• Use the initial velocity to solve for C
a (t )  9.8
v(t )  9.8t  C
160  9.8(0)  C
v(t )  9.8t  160
– Find the height above ground a function of time (t).
• Find the integral of v(t)!
• Use the initial height to solve for C
v(t )  9.8t  160
3  4.9(0)  160(0)  C
h(t )  4.9t 2  160t  3
h(t )  4.9t 2  160t  C
• Ex: A rocket is launched with an initial velocity of 160 m/s from
a platform 3 m above the ground. Assuming the acceleration
due to gravity is a constant -9.8 m/s2, …
– How long was the rocket in the air before it landed?
• Translation: “Find t when h(t) = 0”
• Use a calculator (quadratic formula or graphing) to find the zero!
0  4.9t 2  160t  3
t  32.672 s
– What was the velocity of the rocket when it hit the ground?
v(t )  9.8(32.672)  160
m
 160.18
s
• One way of approximating values using a differential equation
is Euler’s Method
– Euler’s method uses tangent lines and increments of x to get from the
initial value to the desired value
• The drawing below illustrates how Euler’s Method determines f(3) by starting at f(2) = 1
and increasing by increments of .2
• This method will be an under- or overerstimate depending on the concavity of the
graph
(3, f(x))
 dy

f ( xn 1 )  f ( xn )  (x) 

 dx ( xn , yn ) 
(2,1)
2
2.2
2.4
2.6
2.8
3
 dy

f ( xn 1 )  f ( xn )  (x) 
• Euler’s Method:

dx
 ( xn , yn ) 
1. Begin at the initial condition ( the (x, y) given)
2. Find the slope at that (x, y) using the differential equation given
3. Add a tiny increment, or Δx, to x. Also add a tiny increment Δy to y.
This is your new point.
4. Repeat step 2 using the new point. Keep doing this until you get to
the desired value.
dy
 x y
• Ex: Using dx
and f(2) = 0, add increments of
.25 to approximate f(3). dy
 20  2
– First, find slope at (2, 0), dx
–
–
then add .25 to x and (.25)(2) to y to get (2.25, .5)
then find new slope (using (2.25, .5)) and next point…
•
dy
 x y
Ex: Using dx
.25 to approximate f(3).
–
and f(2) = 0, add increments of
…and so on until you reach x = 3. Make a table to help you organize
your data:
(x, y)
dy/dx
Δx
Δy = (dy/dx) Δx
(x+Δx, y+Δy)
(2, 0)
2
.25
.5
(2.25, .5)
(2.25, .5)
2.75
.25
.6875
(2.5, 1.1875)
(2.5, 1.1875)
3.6875
.25
.921875
(2.75, 2.1094)
(2.75, 2.1094)
4.8594
.25
1.2148
(3, 3.3242)
–
–
–
Answer: f(3) ≈ 3.324
Actual f(3) = 4.155
Note that this is an underestimate because the curve is concave up
over [2,3].