# Section 1. 2 - De Anza College

```Differential Equations
Section 1.2: Initial Value Problems (IVPs)
Math 2A
d n ( x)
Solve:
 f ( x, y, y ',..., y ( n 1) ) subject to the conditions
n
dx
y( x0 )  y0
y '( x0 )  y1
…
y ( n 1) ( x0 )  yn 1
Example 1: Given the first order equation y '  y on the interval (, ) . Verify that y  ce x is a oneparameter family of solutions.
If the initial condition is y(0) = 5, then by substituting into y  ce x , we get 5  ce0  c and, therefore,
y  5e x is a solution of the initial value problem y '  y , y(0) = 5.
(The solution passes through the point (0, 5).)
Example 2: Given the 2nd order equation y &quot; y  0 . Verify that y  c1e x  c2e  x is a two-parameter family of
solutions.
The initial conditions are y(1) = 0 and y’(1) = e.
Since y  c1e x  c2e  x
then y '  c1e x  c2e  x
y(1) = 0 and by substitution into y  c1e x  c2e  x , we get 0  c1e1  c2 e 1
y’(1) = e and by substitution into y '  c1e x  c2e  x , we get e  c1e1  c2e 1
If we solve
0  c1e1  c2 e 1 and
e  c1e1  c2e 1 simultaneously by adding, we get
e  2c 1 e1 so
1
c1 
2
If we solve
0  c1e1  c2 e 1 and
e  c1e1  c2e 1 simultaneously by subtracting, we get
e  2c2 e 1 so
1
c2   e 2
2
Therefore,
1
1
y  e x  e 2 e  x and
2
2
1
1
y  e x  e 2 x is a solution of y &quot; y  0 , y(1) = 0, y’(1) = e
2
2
(The solution passes through the point (1, 0) and the slope of the curve at this point is e.)
For initial value problems, does a solution of the problem exist and if a solution does exist, is it unique?
For a first order ODE:
Existence: Does the ODE dy/dx = f(x, y) possess solutions and do any of the solution curves pass through the
point (x0, y0)?
Uniqueness: When can we be certain that there is exactly one solution curve passing through the point
(x0, y0)?
Theorem 1.1 on page 17 gives us the conditions for the existence of a unique solution. We check to see if
f
f(x, y) and
are both continuous in a rectangular region R in the xy-plane.
y
Theorem 1.1: (For 1st order IVPs) Let R be a rectangular region in the xy-plane defined by a ≤ x ≤ b and
f
c ≤ y ≤ d that contains the point (x0, y0) in its interior. If f(x, y) and
are both continuous in a rectangular
y
region R, then there exist some interval I0: x0 – h &lt; x &lt; x0 + h, h &gt; 0 contained in a ≤ x ≤ b and a unique
function y(x), defined on I0, that is a solution of the initial value problem
dy
 f ( x, y ), y ( x0 )  y0
dx
Note: Look at the picture on p. 17.
Example: Determine a region of the plane for which f(x,y) =
dy
 xy has a unique solution whose graph
dx
passes through a point (x0, y0).
f 1

y 2
x
. They are both continuous in any region where x &gt; 0, y &gt; 0 OR
y
dy
 xy will have a unique solution in any region where
x &lt; 0, y &lt; 0. Therefore, the differential equation
dx
x &gt; 0, y &gt; 0 OR x &lt; 0, y &lt; 0 (interval for x is (0, ∞) when y &gt; 0 OR (- ∞, 0) when y &lt; 0).
Since f ( x, y)  xy , then
2 23
Note: A particular solution of the ODE is 2 y  x (You will soon learn how to get this solution!). So, the
3
dy
 xy ,
theorem tells us that there is some interval centered at x = 9 on which the initial value problem
dx
y(9) = 81, has a UNIQUE solution.
1
2
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