Second order homogeneous ODE with real equal roots. Today it’s all about the solution of second-order homogeneous ODE with real equal roots. Let’s start then.

**Second-order homogeneous ODE with real equal roots**

If you are looking for more in second-order ordinary differential equations, do check-in:

**Second-order homogeneous ODE with real and different roots**

**Second-order homogeneous ODE with complex roots**

**What is a particular integral in second-order ODE**

**Exponential functions as particular integrals**

**Trigonometric functions as particular integrals**

**DOWNLOAD READ & PRINT – Second order homogeneous ODE with real equal roots (pdf)**

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 identical.
- Third option: the roots of the auxiliary or the characteristic equation are complex conjugate.

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

So, let me choose the solutions of this equation as

Here is a real number.

Therefore the general solution of the differential equation will be

Here both and are constants.

Now I’ll solve some examples.

**Examples of second-order homogeneous ODE with real equal 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

Therefore I can say .

This shows that the given differential equation has two real equal roots.

Therefore the general solution of the given differential equation is

Hence this is the solution to the given equation.

Next, I’ll show you another example.

**Example 2**

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

**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 rewrite this equation as

Thus I can say .

That shows that the given differential equation has two real equal roots.

Therefore the general solution of the given differential equation is

(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

This gives .

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

This gives

(3)

Next I’ll put and in equation (3).

This means

That means

This gives .

Thus the solution of the differential equation is

Hence this is the solution to the given equation.

Dear friends, this is the end of 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!!

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