# Linear Homogeneous Second- Order Ordinary Differential Equations Analysis and Visualization

Example 10.17 [Repeated Real Characteristic Roots]
Solve and graph the solution of the IVP

Solution: The characteristic equation is

with the repeated real roots

Theorem 10.9 shows that a general solution is given by

Next, calculate values of and to satisfy the initial conditions. Set in Eqn (10.11):

Now, differentiate Eqn. (10.11) with respect to

and set in Eqn. (10.13):

Solve Eqs (10.12) and (10.14) for and

So the solution to the IVP is

End of Example 10.17

Case 3: Complex Conjugate Characteristic Roots - 3 Examples

Example 10.18 [Complex Conjugate Characteristic Roots - Real part Zero ]
Solve the IVP

for arbitrary parameters a and b Then plot the three solutions that correspond to the fol lowing sets of values for a and b .

and plot the results.

Solution: The characteristic equation is it factors as

to get the complex conjugate roots (with real part zero)

Theorem 10.9 shows that a general solution is given by

We calculate values of and to satisfy the initial conditions Differentiate Eqn. (10.16) to obtain

Eqn. (10.16) and (10.17) imply that

Hence the solution to the IVP is

The calculations of and for each of the three sets of initial conditions are left to the reader. Note that all three IC's have the same value
for they just differ in the values for We get

Also observe that all solutions as well as the general solution are periodic with period

End of Example 10.18

Example 10.19 [Complex Conjugate Characteristic Roots - Real Part Positive ]
Solve and graph the solution of the IVP

Solution: The characteristic equation is As this polynomial doesn 't factor readily, use the quadratic formula to calculate the
characteristic roots

Theorem 10.9 shows that a general solution is given by

Next, calculate values of and to satisfy the initial conditions. Set in Eqn (10.18):

Differentiate Eqn (10.18): with respect to

and set in Eqn. (10.20):

Use Eqn. (10.19) to solve Eqn. (10.21) for

So the solution to the IVP is

and with a plot in Figure 10.11.

End of Example 10.19

Example 10.20 [Complex Conjugate Characteristic Roots - Real Part Negative ]
Solve and graph the solution of the IVP

Solution: The ODE is similar to Example 14.15 The characteristic equation is with characteristic roots

Set (the real and imaginary parts of λ ). Then Theorem 10.6 shows that a general solution is given by

Next, calculate values of and to satisfy the initial conditions. Set in Eqn. (10.21):

Differentiate Eqn. (10.21) with respect to

and set in Eqn. (10.23):

Use Eqn. (10.22) to solve Eqn. (10.24) for

So the solution to the IVP is

and with its graph sketched in Figure 10.12

End of Example 10.20

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