Second order homogeneous ODE with complex conjugate roots. Today it’s all about the solution of second order homogeneous ODE with complex conjugate roots.

Let’s start then.

### Second order homogeneous ODE with complex conjugate roots

Now it’s a known thing that the standard form of any second order homogeneous ODE is

Therefore the auxiliary equation for this ODE is

(1)

Next, I’ll solve the equation (1) to get the values of .

So there are three options for the roots of this equation.

- First option: the roots of the auxiliary or the characteristic equation are real and different.
- Second option: the roots of the auxiliary or the characteristic equation are real and equal.
- Third option: the roots of the auxiliary or the characteristic equation are complex conjugate.

In this post, I’ll work with the last option only.

So, let me choose the solutions of this equation as

Therefore the general solution of the differential equation will be

Here both and are constants.

Now I’ll solve some examples.

#### Example of second order homogeneous ODE with complex conjugate roots

Disclaimer: None of these examples is mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here are the examples.

##### Example 1

According to Kreyszig (2005)* “Find the general solution of the following: ”

##### Solution

Here the given second-order differential equation is

Therefore the auxiliary equation (characteristic equation) will be

Now I’ll solve this equation to get the values of .

So I can also rewrite this equation as

This means

So I can say or .

That shows the given differential equation has complex conjugate roots.

Therefore the general solution of the given differential equation is

This is the solution to the given equation.

Next, I’ll show you another example.

##### Example 2

According to Kreyszig (2005)* “Solve the following initial value problem: ”

##### Solution

Here the given second-order differential equation is

Also I know the initial values of and at . That is

My task is to solve this initial value problem.

For that, I’ll start with the general solution of this equation.

###### Step 1

Therefore the auxiliary equation (characteristic equation) will be

Now I’ll solve this equation to get the values of .

Thus the values of will be

Next, I’ll simplify it to get

This means

Now this gives

That shows the given differential equation has complex conjugate roots.

Therefore the general solution of the given differential equation is

(2)

Now I’ll get the values of the constants and .

###### Step 2

Also it is given that

This means at and

So, first I’ll put and in equation (2).

Therefore it will be

That means

Now it is known that and .

Therefore it becomes

This gives .

Now I’ll differentiate equation (2) with respect to .

This gives

(3)

Next I’ll put and in equation (3).

This means

Now I’ll simplify it to get

This gives .

Now I’ll put back and in equation (2).

This will give

Thus the solution of the differential equation is

Hence this is the solution to the given equation.

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

###### Kreyszig, E. (2005): Advanced Engineering Mathematics: International Edition, John Wiley & Sons, 9th Edition, 29th December 2005, Chapter 2, Second-order linear ODEs, p. 59, Problem set 2.2, Q. 1(Example 11), Q. 2 (Example 29).

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