Laplace transform to solve second-order differential equations. Hi guys, today I’ll talk about how to use Laplace transform to solve second-order differential equations. Have a look.

**Laplace transform to solve second-order differential equations**

Now the standard form of any second-order ODE is

Here are constants and is a function of .

In order to solve this equation in the standard way, first of all, I have to solve the homogeneous part of the ODE. After that, I’ll need to get the particular integral for the non-homogeneous part.

If interested, you can read more about it in the category of the second-order ODE.

**But what would happen if I use Laplace transform to solve second-order differential equations**

If I use Laplace transform to solve second-order differential equations, it can be quite a direct approach.

First of all, I don’t need to bother with the homogeneous or non-homogeneous part. It’s all the same.

Secondly, I don’t have to work on initial conditions separately. They will be part of the calculation.

In one of my earlier posts, I have already described in detail how to use Laplace transform to solve the first-order ODE.

For further reading, please check out how to use Laplace transform to solve second-order homogeneous ODE

Now I’ll give an example of how to use Laplace transform to solve second-order differential equations.

**Example of **Laplace** transform to solve second-order differential equations**

*Disclaimer: This example is not mine. I have chosen it from a book. I have also given the due reference at the end of the post.*

So here is the example.

**Example **

According to Stroud and Booth (2011) “Solve the equation 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

(1)

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

In order to get the inverse Laplace transform, first I have to use partial fractions of this expression.

Related post: partial fraction in inverse Laplace transform

As I can see, the factors in the denominator and cannot be reduced further.

So I’ll use the rule for partial fractions of irreducible factors.

Thus the expression will be a sum of two expressions as

(2)

Also, are constants.

Now I’ll find out the values of these constants.

**Step 2**

First of all, I’ll simplify the right-hand side of the equation (2).

Thus it will be

Now I can compare the coefficients on both sides of the numerator.

I will start with the coefficients of .

Thus it will be

(3)

Next, it will be the constants.

So it will look like

(4)

Then I’ll compare the coefficients of to get

Now I’ll substitute the value from equation (4) to this equation to get

(5)

In the end, I’ll compare the coefficients of . Thus it will be

Now from equation (3), I’ll substitute in this equation to get

(6)

Next, I’ll put equation (6) in equation (4) to get

And this brings

Then I can use equation (6) to say

Next, I can get the value of from equation (3) as

At the end, I’ll get the value of from equation (4) as

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

**Step 3**

Thus it will be

Now I’ll simplify it to get

So this means

Therefore from equation (1), I can say that

Next, I’ll use the formulas in Laplace transform to find the value of .

So I can rewrite as

Now, first of all, I’ll find out the value of Then I’ll find out the value of

Then let’s say,

and

(7)

So, now it’s time to get the value of .

**Step 4**

As I can see, the denominator of is Now I can’t factorise it. But I can write it as a sum of two squares like

So this gives

Therefore I can write as

Now I’ll bring in the numerator as well.

Thus it will be

(8)

As I already know, the Laplace transform of is

Also, from the first shift theorem in Laplace transform, I already know that the Laplace transform of is

So in this case, it will be

Again, the Laplace transform of is and together with the first shift theorem the Laplace transform of becomes

Thus will be

Therefore from equation (8), I can say that will be

(9)

Now I’ll find the value of .

**Step 5 **

As I can see, the denominator of is Now I can’t factorise it. But I can write it as a sum of two squares like

Therefore I can write as

So this means

Now I’ll use the formulas in Laplace transform.

As per these formulas, Laplace transform of is Also, the Laplace transform of is

So I can use these two formulas to get the value of as

Thus the value of will be

Now I’ll simplify it to get

(10)

Next, I’ll substitute the values of and from equations (9) and (10) respectively to equation (8) to get the value of as

Then I’ll simplify it to get

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

Dear friends, this is the end of my today’s post on Laplace transform to solve second-order differential equations. 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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