### Differential Equations and Slopefields

```Differential Equations and Slope
Fields
1st period
Differential Equations
• A differential equation is an equation which
involves a function and its derivative
• There are two types of differential equations:
– General solution: is when you solve in terms of y
and there is a constant “C” in the problem
– Particular solution: is when you solve for the
constant “C” and then you plug the “C” into the y
equals equation
Example of a General Solution
Equation
dy
 xy 2
dx
Given:
Step 1: Separation of variables- make sure you
have the same variables together on different
sides of the equation.
1
y
2
dy  xdx
Step 2: Integrate both sides of the equation
1
 y 2 dy   xdx
1
1
  C  x2  C
y
2
1
1 2
 C  x C
y
2
Step 3: Notice how above there is a constant “C”
added to both sides, but you can combine those
constants on one side of the equation and end
up with: 1 1 2

y

2
x C
Step 4: Solve for y
y
1
1 2
x C
2
Example of Particular Solution
dy
 ( x  1)( y  2)
dx
Given:
f(0)= 3
Step 1: Separate the variables like you would
with a general solution equation.
1
dy  ( x  1)dx
y2
Step 2: Integrate both sides of the equation
1
 y  2dy   ( x  1)dx
1
ln y  2  x 2  x  C
2
ln y  2 
1 2
x  xC
2
Step 3: Apply the exponential function to both
sides of the equation
y2  e
Step 4: Solve for y
y  2  eC e
y  2  Ce
y  Ce
1 2
x x
2
1 2
x x
2
1 2
x x
2
2
1 2
x  x C
2
Step 5: Plug in f(0)=3 and solve for C
3  Ce
1 2
(0)  0
2
2
3C2
C 1
Step 6: Plug “C” into the y equals equation you
found in step 4
ye
1 2
x x
2
2
Try Me!
dy
1
1.
 xy
dx
2
2. f(0)=2
3. f(0)=7
dy
 9y
dx
dy
 2(4  y )
dx
4. dy  2 y  0
dx
1.
dy
1

xy
dx
2
1
1
dy 
xdx
y
2
1
1
dy

xdx
y

2
1 2
ln y 
x C
4
ye
1 2
x C
4
y  eC e
y  Ce
1 2
x
4
1 2
x
4
Step 1: Separate the
variables on each side of
the equation
Step 2: Integrate both sides
of the equation
Step 3: Apply the
exponential function to
both sides of the equation
to get y by itself
Step 4: Solve for y to find
the general equation
2. f(0)=2
dy
 9y
dx
1
dy  9 dx
y
1
 y dy   9dx
ln y  9 x  C
y  e9 x  C
y  eC e9 x
y  Ce 9 x
2  Ce 9 ( 0 )
C 2
y  2e 9 x
Step 1: Separation of
variables
Step 2: Integrate both sides
of the equation
Step 3: Apply the
exponential function to
both sides of the equation
Step 4: Plug in values given
and solve for the constant
“C”.
Step 5: Plug in the constant
you found in step 4 into
find the particular
equation
3. f(0)=7
dy
 2(4  y )
dx
1
dy  2dx
4 y
1
 4  y dy   2dx
 ln 4  y  2 x  C
ln 4  y  2 x  C
4  y  e 2 x  C
4  y  e C e 2 x
4  y  Ce 2 x
y  Ce 2 x  4
7  Ce 2(0)  4
7C4
C 3
y  3e 2 x  4
Step 1: Separation of
variables
Step 2: Integrate both sides
of the equation
Step 3: Apply the
exponential function to
both sides
Step 4: Solve for constant
“C” given values f(0)=7
Step 5: Plug constant into y
equals equation to find the
particular equation
4.
dy
 2y  0
dx
1
dy  2dx
y
1
 y dy   2dx
ln y  2 x  C
y  e 2 x  C
y  eC e 2 x
y  Ce
2 x
Step 1: Separate variables
Step 2: Integrate both
sides
Step 3: Apply exponential
function to both sides
Step 4: Solve in terms of y
to find the general
equation
Slope Fields
• Slope fields are a plot of short line segments
with slopes f(x,y) and points (x,y) lie on the
rectangular grid plane
• Slope fields are sometimes referred to as
direction fields or vector fields
• The line segments show the trend of how slope
changes at each point
no slope (0)
undefined
FRQ 2008 AB 5
Consider the differential
x0
dy
y 1

equation dx x 2
where
a. On the axes provided, sketch a slope field for
the given differential equation at the nine
points indicated.
b. Find the particular solution y= f(x) to the
differential equation with initial condition
f(2)=0.
dy
y 1

dx
x2
1
 y  1dy 
1
x

1
C
x
y  1  eC e
y  1  Ce
y  Ce
0  Ce

1
x
1

2
dx
1
C
x
ln y  1  
y 1  e
2


1
x
1
x
1
1
1
C  e 2
y  1 e
x0
(
1 1
 )
2 x
c. For the particular solution y=f(x) described in
part b, find lim f ( x)
x 
lim1  e
x 
1 e
1 1
(  )
2 x
Try Me!
dy
 2x
dx
Consider the differential equation
a. On the axes provided, sketch a slope field for
the given differential equation at the nine
points indicated. Slope = 2
Slope = -2
Slope = 4
FRQ 2004 (FORM B) AB 5
dy
 x 4 ( y  2)
dx
Consider the differential equation
a. On the axes provided, sketch a slope field for
the given differential equation at the twelve
points indicated.
b. While the slope field in part a is drawn at only
twelve points, it is defined at every point in the
xy-plane. Describe all points in the xy-plane
for which the slopes are negative.
x  0 and y  2
c. Find the particular solution y=f(x) to the given
differential equation with the initial condition
f(0)=0.
C.
dy
 x 4 ( y  2)
dx
1
4
dy

x
 y  2  dx
1 5
ln y  2  x  C
5
y  2  eC e
y  2  Ce
y  2  Ce
0  2  Ce
C  2
y  2  2e
1 5
x
5
1 5
x
5
1 5
x
5
1
(0)5
5
1 5
x
5
FRQ 2004 AB 6
dy
 x 2 ( y  1)
dx
Consider the differential equation
a. On the axes provided, sketch a slope field for
the given differential equation at the twelve
points indicated.
b. While the slope field in part a is drawn at only
twelve points, it is defined at every point in the
xy-plane. Describe all points in the xy-plane
for which the slopes are positive.
y  1 and x  0
c. Find the particular solution y=f(x) to the given
differential equation with the initial condition
f(0)=3.
dy
 x 2 ( y  1)
dx
1
2
x
dx


 y  1dy
1 3
x  C  ln y  1
3
e
1 3
x C
3
1 3
x
C 3
e e
Ce
1 3
x
3
1  Ce
1  Ce
 y 1
 y 1
 y 1
1 3
x
3
y
1 3
(0)
3
3
C2
y  1  2e
1 3
x
3
Try Me!
1. f(0)=2
dy
2  6y  4  0
dx
2.
dy
 xe y  0
dx
3.
dy
 y 2 (1  x 2 )
dx
4. f(1)=-1
dy
2x

dx
y
dy
 6y  4  0
dx
dy
 (3 y  2)
dx
1
dy   dx
3y  2
1
 3 y  2dy    dx
2
1.
ln 3 y  2   x  C
3 y  2  e  x C
3y  2  e e
C
x
3 y  2  Ce  x
3 y  Ce  x  2
1
2
Ce  x 
3
3
1
2
2  ( Ce 0  )
3
3
8
1
 C
3
3
C 8
y
y
8 x
2
e 
3
3
2.
dy
 xe y  0
dx
e  y dy  xdx
y
e
 dy   xdx
1 2
e  x  C
2
1 2
 y  ln  x  C
2
y
1 2
y   ln  x  C
2
3. dy
 y (1  x )
2
2
dx
1
2
dy

(1

x
)dx
2
y
1
2
dy

(1

x
)dx
 y2 
1 3
1
y  x  x C
3
1 3
1
y  x  xC
3
1
y
1 3
x  xC
3
4. f(1)=-1 dy
2x
dx
y
ydy  2 xdx

 ydy   2 xdx
1 2
y   x2  C
2
y 2  2 x 2  C
(1) 2  2(1) 2  C
1  2  C
C 3
y 2  2 x 2  3
y   2 x  3
Slope Field Example
x
y
y’ = x + y
-1
-1
-2
-1
0
-1
-1
1
0
0
0
0
1
-1
0
1
0
1
1
1
2
FRQ 2006 AB 5
dy 1  y

dx
x
Consider the differential equation
where x  0
a. On the axes provided, sketch a slope field for
the given differential equation at the eight
points indicated.
b. Find the particular solution y=f(x) to the
differential equation with the initial condition
f(-1)=1 and state its domain.
dy 1  y

dx
x
1
1
 xdx   1  y dy
ln x  C  ln 1  y
Cx  1  y
y  C x 1
1  C 1  1
C2
y  2 x  1
The domain is x<0
Review!
x
y
y’=4x/y
-1
-1
-1
0
-1
0
1
-1
4
Und.
-4
0
0
0
1
0
1
-1
Und.
0
-4
1
1
0
1
Und.
4
Review Continued…
dy
y
x
dx
ydy  xdx
 ydy   xdx
1 2 1 2
y  x C
2
2
y2  x2  C
y   x2  C
Step 1: Separation of
variables
Step 2: Integrate both
sides of the equation
Step 3: Solve in terms
of y to find the
general solution
Review Continued…
f(1)=0
dy
 xe  y
dx
e y dy  xdx
e
y
dy   xdx
1 2
x C
2
1
y  ln x 2  C
2
ey 
0  ln
1 2
(1)  C
2
1
C
2
1
C
2
1
1
y  ln x 2 
2
2
1
Step 1: Separate the
variables
Step 2: Integrate both
sides
Step 3: Plug in f(1)=0 to
find the constant “C”
Step 4: Plug the constant
you just found into the y
equals equation
Bibliography
• http://www.math.buffalo.edu/~apeleg/mth30
6g_slope_field_1.gif