Transcript Initial Value Problems, Slope Fields

Initial Value
Problems,
Slope Fields
Section 6.1a
Do Now: #26 on p.313
Determine which graph shows the solution of the given initial
value problem without actually solving the problem.
dy
 x
dx
y  1  1
The correct graph:
(–1,1)
Can you explain
why this is the
correct graph???
An equation like
dy
 y ln x
dx
containing a derivative is a differential equation.
Initial Value Problem – the problem of finding a function
y of x when given its derivative and its value at a particular
point.
Initial Condition – the value of f for one value of x.
Solution to the Differential Equation – all of the
functions y that satisfy the differential equation.
Solution to the Initial Value Problem – a particular
solution that fulfills the initial condition.
An equation like
dy
 y ln x
dx
containing a derivative is a differential equation.
Differential equations from previous chapters???
 Implicit differentiation yields differential
equations
 Related rates equations are differential
equations
First New Example
Suppose $100 is invested in an account that pays 5.6%
interest compounded continuously. Find a formula for the
amount in the account at any time t.
Let t = 0 when the initial $100 is deposited in the account
We can model this situation with the initial value problem:
dy
 0.056 y
dx
y  0   100
First New Example
Suppose $100 is invested in an account that pays 5.6%
interest compounded continuously. Find a formula for the
amount in the account at any time t.
dy
 0.056 y
dx
We need a function whose derivative is a
constant multiple of itself…
 Exponential functions have this property!
…because:
y  t   Ce
0.056t
dy
0.056 t
0.056 t
 C  0.056 e
 0.056  Ce


dt
 0.056 y  t 
First New Example
Suppose $100 is invested in an account that pays 5.6%
interest compounded continuously. Find a formula for the
amount in the account at any time t.
Now, apply the initial condition:
y  t   Ce
y  0   100
0.056t
 0.056 0
100  Ce
100  C
In this case,
y  t   100e
0.056t
First New Example
Suppose $100 is invested in an account that pays 5.6%
interest compounded continuously. Find a formula for the
amount in the account at any time t.
In general, each member of the family of functions
y  t   Ce
0.056t
is a solution of the differential equation
dy
 0.056 y
dx
Let’s learn how to “see” this
family of functions…
Definition: Slope Field (Direction Field)
A slope field or direction field for the first order
differential equation
dy
 f  x, y 
dx
is a plot of short line segments with slopes
for a lattice of points x, y in the plane.


f  x, y 
They are useful for “seeing” solutions to
differential equations even when explicit
solutions are difficult to come by…
Guided Practice
Plot the solution curves of the differential equation
2 xy
y  
2
1 x
Can we find the original function???
Let’s use a new calculator program to see the slope field,
as well as some specific solution curves…
Guided Practice
Solve the given initial value problem. Support your answer
by overlaying your solution on a slope field for the differential
equation.
dy 1
 2 x
dx x
y  2  1
2
x
dy
1
2

y


x


C
dx

x

x
dx


 dx 
2
Initial Condition:
2
2 1
1  C
2 2
1
C 
2
Slope Field and Graph?
2
x 1 1
Solution: y 
 
2 x 2
Guided Practice
Solve the given initial value problem.
dy
 cos x  sin x
dx
y    1
dy
 dx dx    cos x  sin x  dx
y  sin x  cos x  C
1  sin   cos   C
1  0   1  C
Solution:
C 0
y  sin x  cos x
Guided Practice
Solve the given initial value problem.
y e   0
dy 1

dx x
3
dy
1
dx

dx


 dx   x 
y  ln x  C
0  ln  e   C
0  3  C  C  3
3
Solution:
y  ln x  3
Guided Practice
Solve the given initial value problem.
2
d y
 2  6x
2
dx
y  0   1 y  0   4
2
d y
 dx2 dx    2  6 x  dx
First Derivative:
dy
2
2
 2 x  3x  C1  2 x  3 x  4
dx
4  2  0   3  0   C1  C1  4
2
Guided Practice
Solve the given initial value problem.
2
d y

y
0

4
y
0

1

2

6
x




2
dx
dy
2
dx

2
x

3
x

4
dx


 dx 
y  x  x  4 x  C2
2
3
1  0  0  4  0  C2  C2  1
2
Solution:
3
y   x  x  4x  1
3
2
Guided Practice
Use the given information about a body to find the body’s
position s at time t.
a  cos t
s  0  1
v  0  1
dv
 dt dt    cos t  dt v  sin t  C1
1  sin 0  C1 C1  1
ds
 dt dt    sin t  1 dt s   cos t  t  C2
1   cos 0  0  C2 C2  2
s   cos t  t  2