How to Solve Differential Equations Using Laplace Transform

TL;DR

To solve a differential equation using the Laplace Transform, first apply the transform to the initial value problem, isolating Y(s). Then, perform the inverse Laplace Transform to find the solution, yielding y(t) = −e^(−3t) + 3te^(−3t) for given conditions y(0) = 1 and y′(0) = 6.

Transcript

okay we're going to solve this initial for problem by using the laas transform so let's go ahead and take the laas right here right and this is the original laas transform so here we go we first have the second derivative and you have to remember that this is going to give us s² and then we will have y of s right and this is supposed to be a capita... Read More

Key Insights

• 🔨 Laplace Transform is a powerful tool for solving initial value problems.
• ❓ The process involves transforming the equation, isolating the variable, and finding the inverse Laplace Transform to obtain the solution.
• 🧑‍🏭 Factoring out the Laplace Transform variable helps in simplifying the equation.

Questions & Answers

Q: How do you solve the Section 7.5 #3 differential equation with the Laplace transform?

Apply the Laplace transform to every term, substitute y(0) = 1 and y′(0) = 6, and collect all terms containing Y(s). This produces Y(s) = −s/(s² + 6s + 9), whose inverse Laplace transform is y(t) = −e^(−3t) + 3te^(−3t).

Q: How are the initial conditions used in the transformed equation?

For the second derivative, the transformation includes −s y(0) − y′(0), so the values 1 and 6 contribute −s and −6. The transformed first-derivative term also uses y(0) = 1, producing the constant term that later cancels the −6.

Q: How is Y(s) isolated after applying the Laplace transform?

Collect the three terms containing Y(s) and factor it out to obtain (s² + 6s + 9)Y(s). The remaining constant terms −6 and +6 cancel, and moving s to the other side gives Y(s) = −s/(s² + 6s + 9).

Q: Why is the denominator written as (s + 3)²?

The quadratic s² + 6s + 9 factors as (s + 3)². This form makes it easier to match the transformed expression with inverse Laplace transform patterns.

Q: Is partial fraction decomposition required for this problem?

No, the example presents partial fractions as an available choice but uses a quicker rearrangement. Adding and subtracting 3 in the numerator creates terms that match 1/(s + 3) and 1/(s + 3)².

Q: How is the numerator rearranged for the inverse Laplace transform?

After moving the leading negative sign outside, the numerator s is rewritten using s + 3 − 3. This separates the expression into terms involving 1/(s + 3) and 3/(s + 3)² while preserving the overall sign.

Q: What inverse Laplace transform pairs are used in the solution?

The term 1/(s + 3) becomes e^(−3t). The term 1/(s + 3)² becomes te^(−3t), with its coefficient and signs producing the second term 3te^(−3t).

Q: What is the final solution to the initial value problem?

The final solution is y(t) = −e^(−3t) + 3te^(−3t). It results from applying the inverse Laplace transform to Y(s) = −s/(s + 3)².

Summary & Key Takeaways

• The video demonstrates the process of using Laplace Transform to solve an initial value problem.

• It breaks down the steps involved in transforming the equation and isolating the variable.

• The final solution is obtained by performing the inverse Laplace Transform on the transformed equation.