Here,

Therefore:

Integrating:

Applying the limits:

For s > 0:

Hence:

Here:

Therefore:

Since 1! = 1:

Put n = 2:

Since:

Therefore:

Now:

Therefore:

Using the definition:

Combine the exponential terms:

Therefore:

Start with:

Replace a by −a:

Therefore:

The standard Laplace transform of sin(at) is:

For example, if a = 3:

Using the standard Laplace transform:

For example:

Use the identity:

Taking Laplace transform:

Taking common denominator:

Use:

Taking Laplace transform:

After simplifying:

Given:

Therefore:

Using the constant multiplication property:

Since L{f(t)} = F(s):

Using linearity:

Now:

Therefore:

We know:

Therefore:

Also:

Hence:

Here a = 2.

Multiplying by 5:

Use:

Here a = 2:

Multiply by 3:

For a = 3:

Multiply by 4:

Since:

Therefore:

We know:

Therefore:

We know:

Therefore:

By the derivative property of the Laplace transform:

Since:

we get:

The initial value f(0) is therefore included automatically in the transformed derivative.

We know:

Apply the same derivative rule again:

Substitute:

Expand:

From the second derivative:

Now take the transform of the third derivative:

Substitute the previous result:

Expand:

Given:

Using the formula:

Substitute f(0) = 2:

Given:

Substitute into the derivative formula:

Therefore:

Suppose:

Differentiate F(s) with respect to s:

Therefore:

For n = 2:

Since (−1)² = 1:

We know:

Therefore:

Differentiate:

Since L{tf(t)} = −F′(s):

Let:

Differentiate with respect to s:

Using L{tf(t)} = −F′(s):

We know:

Differentiate:

Therefore:

Hence:

Multiplying f(t) by e^at shifts the s-variable.

Therefore:

Use:

Replace a by −a:

Therefore:

Using the standard result:

Here a = 2:

The general result is:

For n = 2:

Since 2! = 2:

Since:

Therefore:

We know:

Here a = 2, so replace s by s−2:

We know:

Apply the exponential shifting property:

Therefore:

Start with:

Since the exponential is e^−3t, replace s by s+3:

We know:

Because of e^−2t, replace s by s+2:

Since:

Taking inverse Laplace transform:

We know:

Therefore:

We know:

Therefore:

Using:

Therefore:

Comparing with the standard transform pair:

Hence:

We know:

Therefore:

Compare the given expression with the standard transform:

Hence:

Therefore:

Compare the expression:

with the standard Laplace transform of sinh(at).

Compare the given expression with:

Therefore:

Use partial fractions:

Putting s = 0 gives A = 1.

Putting s = −1 gives B = −1.

Taking inverse Laplace:

Taking inverse Laplace:

Therefore:

Factor the denominator:

Write:

Solving gives A = 1 and B = 1.

Hence:

Taking inverse Laplace:

Compare with:

Here a² = 4, so a = 2.

For a = 2:

Hence the inverse is sin(2t).

Therefore:

Use:

Here a = 3.

Putting a = 3:

Use:

For a = 2:

Use:

For a = 3:

Use:

Here a = −1:

Therefore:

We know:

Apply the shifting rule with s replaced by s+2:

Start with:

Replace s with s+2:

The standard form is:

Comparing:

Therefore:

Use:

Here a = 1 and b = 2.

Compare with:

Here:

Therefore a = −3.

Also b² = 16, so b = 4.

Use:

Here a = −3 and b = 4.

Use:

Comparing:

Thus a = −1.

Also b = 3.

The Laplace transform is linear:

Therefore:

A constant multiplier can be taken outside the Laplace transform:

Thus:

Using the definition:

The entire integrand is zero.

This is one of the fundamental Laplace transform formulas.

For example, n = 3:

Use:

For a = 1:

Put a = 1:

Use:

For a = 1:

For a = 1:

Use:

Here a = −1:

Therefore:

Putting a = 2:

Multiplying by 4:

Use linearity:

Therefore:

Multiply by 3:

Multiply by 5:

Now:

Therefore:

A standard Laplace property states:

This property is useful when the time-domain function is divided by t.

The derivative property converts derivatives in the time domain into algebraic expressions involving s in the Laplace domain.

This conversion makes many differential equations easier to solve algebraically.

Use:

Given:

and f(0)=1.

Substitute:

Use partial fractions:

Solving the coefficients gives:

Therefore:

Taking inverse Laplace:

Complete the square:

Rewrite the numerator:

Therefore:

Now:

Also:

We know:

Let:

Differentiate:

Using L{tf(t)} = −F′(s):

We know:

Differentiate with respect to s:

Hence:

Therefore the correct expression is:

Start with:

Differentiate:

Since:

we get:

First derivative:

Apply the derivative property once more:

Substitute the first derivative result:

Expand:

This formula is especially important when solving second-order differential equations using Laplace transforms.