Example
10.12:
Solve the IVP
Solution: We saw in Example 10.6 the general solution is
Calculate values of 
and
to satisfy the initial conditions. Let in Eqn. (10.6):
or
We need another relationship between
and 
in order de termine each of them
uniquely. Since we are also given a value for the derivative
we can hope to obtain another relationship by differentiating Eqn. (10.6) with
respect to 
Let 
in Eqn. (10.7):
or
Solve for 
and
:
So the solution to the IVP is
End
of Example 10.12
The analysis in Example 10.10 implies that we can determine
and 
for any set of
initial conditions 
Example
10.13
Solve the IVP
Solution: Write
The initial conditions imply
Now solve this system for
and 
Rewrite this system of equations as
First add the two equations , then subtract the two equations to get
Hence
[You should check that these formulas give the same results using the initial
data in Example 10.11.]
End
of Example 10.13
10.5 VISUALIZATION OF SOLUTIONS
Case 1: Distinct Real Characteristic Roots - 3 Examples
Example
10.14 [Distinct Real Characteristic Roots ]
Solve and graph the solution of the IVP
Solution: The characteristic equation is
Factor as
to get the distinct real roots
Theorem 10.9 shows that the general solution is given by
Next, calculate values of
to satisfy the initial conditions. Set
in Eqn.
(10.8):
Differentiate Eqn. (10.9) with respect to
:
and set 
in Eqn. (10.10):
Solve Eqns(10.9) and (10.11) for
:
So the solution to the IVP is
and is plotted be low in Figure 10.6.
End
of Example 10.14
Example
EXAMPLE 10.15 [Distinct Real Characteristic Roots]
Solve ODE
for each of the initial conditions
and plot the results.
Solution: The characteristic equation is
As this polynomial doesn 't factor
readily, use the quadratic formula to calculate
the characteristic roots
to get the distinct real roots
Theorem 10.9 shows that the general solution is given by
The calculations for each of the three initial conditions's are left to the
reader. Note that all three sets of initial conditions have the same value
for 
they just differ in the values for
We get
The important thing to gt from this example is the different behaviors of the
solutions as illustrated in Figure 10.7. Observe how the steepness
of the solutions at 
increase with increasing values of
End
of Example 10.15
Example
10.16 [Distinct Real Characteristic Roots]
Solve same ODE as in Example 10.14
for each of the initial conditions
and plot the results.
Solution: The general solution is the same as in Example 10.13, namely
We leave it to the reader to calculate the coefficients
and 
for each of the four
sets of initial conditions. The important thing to gt from
this example is the different behaviors of the solutions as illustrated in
Figure 10.8. Observe how the steepness of the solutions at
increase with increasing values of 
End
of Example 10.16
Case 2: Repeated Real Characteristic Roots - 1 Example
