Laplace transform to solve second-order homogeneous ODE. Hi guys, today I’ll talk about how to use Laplace transform to solve second-order homogeneous ODE.

**Laplace transform to solve second-order homogeneous ODE**

Now the standard form of any second-order homogeneous ODE is

Here are constants.

In order to solve this equation in the standard way, first of all, I have to write the auxiliary equation.

Then I’ll solve the equation. Next, I have to check the nature of the solutions of the auxiliary equation to get the general solution of the ODE.

For example, if the roots are real and different, the general solution will be of one type.

If the roots are real and equal, the general solution will be different than the previous one.

And if the roots are complex conjugate numbers, the general solution will be altogether different.

Now if I know the initial values of and , then I can put these values in the general solution to solve the initial value problem.

So, as one can see, it’s quite a process!

**But what would happen if I use Laplace **transform** to solve it?**

If I use Laplace transform to solve this ODE, it can be quite a direct approach.

If interested, please check out how to use Laplace transform to solve the first-order ODE.

So, with Laplace transform, the given ODE will be

Now Laplace transform of is .

Thus it will be

Now Laplace transform of is

Here is the initial value of . And is the initial value of .

Similarly, Laplace transform of is

Also, Laplace transform of is

Therefore Laplace transform of the equation will be

Now I’ll simplify it to get

Therefore the value of will be the inverse Laplace transform of .

So this means

Now I’ll give some examples on that.

**Examples of Laplace transform to solve the second-order homogeneous ODE**

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)* “Solve the following initial value problem by Laplace transform:

”

**Solution**

Here the given differential equation with initial condition is

Now I have to solve this equation using Laplace transform.

**Step 1**

So it will be

This means

Now I already know from the given equation that .

Therefore I’ll simplify this equation to get

Thus the value of will be

Now I’ll use partial fractions to get the inverse Laplace transform of this expression.

So I’ll rewrite the expression as a sum of two fractions like

If you want to know how could I write in this way, please read more at

partial fractions of equal degree expressions

Thus my task is now to find the values of and .

**Step 2**

Now here the denominators on the right-hand side have only , not or any other higher power of . So I can use the cover-up rule to find the inverse Laplace transform.

As I can see, the value of is the coefficient of .

So, by cover up rule, it will be

In the same way, now I’ll get the value of .

So here the value of is the coefficient of .

Thus I will use the same method to get the value of as

Therefore I can say

Now I’ll find out the value of as

So I can rewrite as

Next, I’ll use formulas in Laplace transform to get the inverse transformation of this expression.

So this gives

Hence I can conclude that is the solution to this example.

Now I’ll give another example.

**Example 2**

According to Kreyszig (2005)* “Solve the following initial value problem by Laplace transform:

”

**Solution**

Here the given differential equation with initial condition is

Now I have to solve this equation using Laplace transform.

**Step 1**

So it will be

This means

Now I already know from the given equation that .

Therefore I’ll simplify this equation to get

Thus the value of will be

So my next step is to find the inverse Laplace transform.

**Step 2**

As I can see here that the denominator cannot be factorised. But I can rewrite it as a sum of two squares like

Next, I’ll also bring the term in the numerator like

(1)

Now, from the standard formulas in Laplace transform, I already know that

So this means for ,

Again, from the first shift theorem in Laplace transform, I also know that

Thus for ,

Hence it means

(2)

(3)

Now I’ll put back equations (2) and (3) to equation (1) to get

Hence my conclusion is is the solution to this example.

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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