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

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

In my earlier posts on the first-order ordinary differential equations, I have already shown how to solve these equations using different methods. These are

- separation of variables,
- transformation of variables,
- substitution of variables,
- integrating factor and
- Bernoulli’s equations.

Today I’ll show how to use Laplace transform to solve these equations.

If I use Laplace transform to solve differential equations, I’ll have a few advantages. These are

- only one method for first-, second- or higher-order differential equations.
- initial conditions will be a part of the calculation.

So I’ll give a simple example now.

Suppose I have a first-order ODE like

Here are constants. Also, I have the initial value of as

Next, I’ll use the Laplace transform to solve this equation. So it will be

This means

Now Laplace transform of is

Here is the initial value of . Similarly, Laplace transform of is

Also, Laplace transform of any constant is . Therefore Laplace transform of the equation will be

Now I’ll replace with 1. Then I’ll simplify it to get

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

So this means

Next, I have to get the inverse Laplace transform of this term to get the solution of the differential equation.

So, this is the basic method.

Now I’ll give some examples of how to use Laplace transform to solve first-order differential equations.

**Examples of Laplace transform to solve first-order differential equations**

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.

So it will be

This means

Now I already know from the given equation that .

Therefore I’ll simplify this equation to get

Next, I’ll find out the value of as

So I’ll use one of the standard formulas in Laplace transform.

Thus it will be

Hence I can conclude that the solution of the differential equation will be

This is the answer to this example.

Now I’ll give you 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.

First of all, I’ll apply the Laplace transform to this equation.

**Step 1**

So it will be

This means

Also I already know from the given equation that .

Therefore I’ll simplify this equation to get

Next, I’ll simplify it further so that I can write in terms of .

**Step 2**

Thus it will be

So this gives

Therefore 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 irreducible quadratic factors

partial fractions of equal degree expressions

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

**Step 3**

At first, I’ll simplify the right-hand side of the expression as

Now on both sides the denominator is the same. So I can say the numerator on both sides is equal, that is,

Next, I’ll compare the coefficients on both sides.

First of all, I’ll compare the coefficients of .

Thus it will be

(1)

Then I’ll compare the coefficients of .

Thus it will be

From equation (1), I can say that

(2)

At last, I’ll compare the constants on both sides.

Thus it will be

From equation (2), I can say that

(3)

Now I’ll substitute the value of from equation (3) to equation (2) to get

Next, I’ll substitute the value of from equation (3) to equation (1) to get

Thus I can say

Now I’ll find out the value of as

(4)

**Step 4**

So I can rewrite the equation (4) as

Again I can rewrite it as

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

and

Thus the value of will be

Hence I can conclude that

is the solution of this eample.

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!!

Sachin says

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Jose Luis Castro says

What’s the relation with integrating factor method?

Dr. Aspriha Peters says

None. Please read it carefully and you will see it.

Thanks.

Mahzeb says

Its really helpful. Clear explanation for the solution of DE’s by Laplace transform

Dr. Aspriha Peters says

Hi Mahzeb,

Thanks a lot.

Regards,