Use the Laplace transform to solve the given initial-value problem. y^''+9 y=e^t, y(0)=0, y^

Use the Laplace transform to solve the given initial-value...

Key Concepts

Laplace Transform

The Laplace transform is a powerful integral transform used to convert functions of time into functions of a complex variable. This tool simplifies differential equations by transforming them into algebraic equations, making them easier to solve, especially when initial conditions are provided.

Initial-Value Problems

Initial-value problems involve differential equations paired with specific values for the function and its derivatives at a starting point. These conditions allow one to determine constants in the solution when using methods like the Laplace transform, which naturally incorporates these conditions early in the solution process.

Transformation of Differential Equations

Applying the Laplace transform to differential equations converts derivatives into algebraic expressions. For second-order ordinary differential equations, each differentiation is replaced by terms involving the Laplace variable and the initial conditions, reducing the problem to an algebraic equation that can be manipulated and solved.

Inverse Laplace Transform

Once the algebraic equation in the Laplace domain has been solved for the transformed function, the inverse Laplace transform is used to convert the solution back into the time domain. Techniques such as partial fraction decomposition are often employed to facilitate this step by breaking down complex rational expressions into simpler fractions that match standard inverse transform forms.

Partial Fraction Decomposition

This algebraic method is used to simplify complex rational expressions into a sum of simpler fractions. In the context of inverse Laplace transforms, partial fraction decomposition enables the application of standard Laplace transform pairs, making it easier to identify and apply the inverse transform to reach the final solution in the time domain.

Transcript

00:01 In this question, we are required to solve the initial value problem.

00:05 Y double dash plus 9y is equal to e to the power t by using the laplace transform.

00:13 We have y0 is equal to 0 and y dash 0 is equal to 0.

00:20 So let's see how to solve this question.

00:23 First of all, let's take the laplace transform of each of the term.

00:27 So we can write laplace transform of y double dash plus 9 into laplace transform of y is equal to laplace transform of e to the power t.

00:40 We know that the laplace transform of y double dash is equals to s square ys minus s y -0 minus y dash 0.

00:57 And we know laplace transform of y is equal to y s and the laplace transform of e to the power a t is equal to 1 upon s minus a.

01:14 Now apply all these formulas so we can write s square ys minus s y0 minus y -0 plus 9 into ys is equal to and the laplace transform of e to the power t will be equal to 1 upon s minus 1.

01:42 Now substitute the values of y 0 and y -dash 0 so we will have s square ys plus 9 y is equal to 1 upon s minus 1.

02:00 Now from the these two terms let's take out ys as a common, so we will have s square plus 9 into ys is equals to 1 upon s minus 1.

02:13 Therefore, ys will be equal to 1 upon s minus 1 into s square plus 9.

02:23 Now to further simplify this expression, let's apply the partial fraction decomposition technique...

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