How to use partial fractions in inverse Laplace transform. Today it’s all about the partial fractions in inverse Laplace transform.

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### Partial fractions in inverse Laplace transform

#### Examples of partial fractions in inverse Laplace transform

Disclaimer: These examples do not belong to me. I have chosen these from a book. At the end of the post, I have given the due reference.

So here are the examples of how to use partial fractions in inverse Laplace transform.

##### Example 1

According to Stroud and Booth (2011) “Determine the inverse transformation of the following: ”

##### Solution

Here the given expression is

So I can see that in the denominator, the factor is repeated thrice.

In order to get the partial fraction of the expression, I’ll use the same method as the partial fraction of expressions with repeated roots. So this will be my first step.

###### Step 1

Therefore I’ll rewrite the expression as

Now I’ll simplify the right-hand side.

Thus it will be

So this means

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

I will start with the coefficient of .

Thus it will be

Next, it will be the coefficient of .

So it will look like

In the end, it will be the constants on both sides.

Thus it will look like

So the expression in the partial fraction form looks like

Now I’ll do the inverse Laplace transform of this expression.

###### Step 2

So the inverse Laplace transform of this expression will be

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

Also, I already know from the Laplace transform of any function multiplied by a variable that

and

.

Therefore I can say that

and

So the inverse Laplace transform of this expression will be

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

Now I’ll go to my next example on how to use partial fractions in inverse Laplace transform.

##### Example 2

According to Stroud and Booth (2011) “Determine the inverse transformation of the following: ”

##### Solution

Here the given expression is

So I can see that in the denominator, none of the factors can be reduced further.

In order to get the partial fraction of the expression, I’ll use the same method as the partial fractions of irreducible quadratic factors. So this will be my first step.

###### Step 1

Therefore I’ll rewrite the expression as

Now I’ll simplify the right-hand side.

Thus it will be

So this means

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

I will start with the coefficient of .

(1)

Next, it will be the coefficient of .

(2)

Now it’s the time for the coefficient of .

(3)

In the end, it will be the constants on both sides.

(4)

Now I’ll solve equation (1) – (4) to get the values of and .

So from equation (1), I can say

Now I can substitute this value in equation (3) to get

Then the value of will be

Again, from equation (2), I can say

Now I can substitute this value in equation (4) to get

So the expression in the partial fraction form looks like

Now I’ll do the inverse Laplace transform of this expression.

###### Step 2

So the inverse Laplace transform of this expression will be

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

Similarly,

So I can say that

Hence I can conclude that

is the solution to this example.

Now I’ll go to my next example on how to use partial fractions in inverse Laplace transform.

##### Example 3

According to Stroud and Booth (2011) “Determine the inverse transformation of the following: ”

##### Solution

Here the given expression is

So I can see that in the denominator, none of the factors can be reduced further.

In order to get the partial fraction of the expression, I’ll use the same method as in example 2. So this will be my first step.

###### Step 1

Therefore I’ll rewrite the expression as

Now I’ll simplify the right-hand side.

Thus it will be

So this means

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

I will start with the coefficient of .

(5)

Next, it will be the coefficient of .

(6)

In the end, it will be the constants on both sides.

(7)

Now from equation (5), I can say

Again from equation (7), I can say

So this means

Next, I’ll substitute these values of in (6) to get

Then the values of and will be

and

So the expression in the partial fraction form looks like

Now I’ll do the inverse Laplace transform of this expression.

###### Step 2

So the inverse Laplace transform of this expression will be

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

So I can say that

Next, I’ll find out the inverse Laplace transform of .

Now, for that, I’ll rewrite the expression as

So this means,

Thus I can say the inverse Laplace transform of this expression will be

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

and

And, as per the first shift theorem in Laplace transform,

and

So I can say

Now this gives

Thus it means \mathcal{L}^{-1}\left\{ \frac{2 – 11s}{(s – 2)(s^2 + 2s + 2)} \right\} = -2 e^{2t} + 2e^{-t}\cos t – 5 e^{-t}\sin t.

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

Dear friends, this is the end of my today’s post on how to use partial fractions in inverse Laplace transform. 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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