Shifted data problem. Hello friends, today I’ll talk about the shifted data problem in Laplace transform. Have a look!!

If you are looking for more in how to use Laplace transform to solve differential equations, do check-in:

**Laplace transform in the first-order ODE**

**How to use Laplace transform to solve a second-order homogeneous ODE**

**How to use Laplace transform to solve a second-order ODE**

Shifted data problem_compressed

### Shifted data problem

Now let’s suppose I have got a differential equation like . And I have to solve this equation using Laplace transform.

As I can see, here the initial condition is at . So this means, in this example, the initial condition is at some but not at . Hence I can set so that gives . Thus I can use Laplace transform to solve this equation.

So this is the core of the shifted data problems in Laplace transform. Now I’ll solve some examples for you.

**Solved examples in shifted data problem**

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**

As I have mentioned above, here . So I can set . Then the Laplace transform of the given equaiton 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 .

**Step 2**

Hence it will be

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

Thus it will be

As I already know, . So 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.

**Step 1**

As I have mentioned above, here . So I can set . Then the Laplace transform of the given equaiton will be

This means

Now I already know from the given equation that and .

Therefore I’ll simplify this equation to get

Now I’ll factorise as

Thus I can say

Next, I’ll find out the value of .

Hence it 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

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

So this gives

As I already know, . So it will be

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

This is the answer 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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