Solving Differential Equations With Laplace Transform

by TextBrain Team 54 views

Introduction

Hey guys! Today, we're diving into the fascinating world of differential equations and how to solve them using the powerful Laplace transform technique. Differential equations are mathematical expressions that describe the relationship between a function and its derivatives. They pop up everywhere in science and engineering, from modeling the motion of objects to understanding the flow of heat. The Laplace transform is a fantastic tool that converts differential equations into algebraic equations, making them much easier to solve. So, let's get started and see how we can tackle this problem step-by-step!

Problem Statement

Our mission, should we choose to accept it (and we do!), is to solve the following second-order linear differential equation:

y'' + 5y' - 6y = 21e^t

with the initial conditions:

  • y(0) = -1
  • y'(0) = 9

These initial conditions are crucial because they give us specific values of the function and its derivative at a particular point, which helps us nail down a unique solution. Without them, we'd have a whole family of solutions instead of just one.

Applying the Laplace Transform

So, how do we use the Laplace transform? The key idea is to transform the differential equation from the time domain (where our variable is t) to the frequency domain (where our variable is s). This transformation turns derivatives into algebraic expressions, simplifying the problem. Let's break it down:

  1. Transform each term: We apply the Laplace transform to each term in the differential equation. Remember these important Laplace transforms:

    • L{y''(t)} = s^2Y(s) - sy(0) - y'(0)
    • L{y'(t)} = sY(s) - y(0)
    • L{y(t)} = Y(s)
    • L{e^{at}} = 1/(s-a)

    Where Y(s) is the Laplace transform of y(t). These are the bread and butter of solving differential equations with Laplace transforms, so make sure you're comfy with them. You might want to keep these formulas handy as we work through the steps, or even better, try to memorize them – they'll be your trusty companions in this mathematical journey!

  2. Substitute initial conditions: We plug in the given initial conditions y(0) = -1 and y'(0) = 9 into the transformed equation. This is where those initial conditions really shine, helping us move from a general solution to a specific one.

  3. Transforming the equation: Applying the Laplace transform to our differential equation:

    L{y'' + 5y' - 6y} = L{21e^t}

    This gives us:

    [s^2Y(s) - sy(0) - y'(0)] + 5[sY(s) - y(0)] - 6Y(s) = 21/(s-1)

    Now, we substitute the initial conditions y(0) = -1 and y'(0) = 9:

    [s^2Y(s) - s(-1) - 9] + 5[sY(s) - (-1)] - 6Y(s) = 21/(s-1)

  4. Simplify: Now we simplify the equation by expanding and combining like terms. This step is all about tidying up our equation so it's easier to work with. Trust me, a clean equation is a happy equation!

    (s^2 + 5s - 6)Y(s) + s + 5 - 9 = 21/(s-1)

    (s^2 + 5s - 6)Y(s) + s - 4 = 21/(s-1)

Solving for Y(s)

The next step is to isolate Y(s), which represents the Laplace transform of our solution. This involves a bit of algebraic manipulation, but nothing we can't handle! It's like untangling a knot, one step at a time.

  1. Move terms: We move the terms that don't involve Y(s) to the right side of the equation:

    (s^2 + 5s - 6)Y(s) = 21/(s-1) - s + 4

  2. Combine terms: We find a common denominator and combine the terms on the right side:

    (s^2 + 5s - 6)Y(s) = [21 - (s - 4)(s - 1)] / (s - 1)

  3. Expand and simplify the numerator: Let's expand the numerator. Remember to distribute carefully and watch out for those pesky negative signs. Every sign matters!

    (s^2 + 5s - 6)Y(s) = [21 - (s^2 - 5s + 4)] / (s - 1)

    (s^2 + 5s - 6)Y(s) = (-s^2 + 5s + 17) / (s - 1)

  4. Factor and isolate: Now, we factor the quadratic term on the left side and divide both sides to isolate Y(s):

    ((s + 6)(s - 1))Y(s) = (-s^2 + 5s + 17) / (s - 1)

    Y(s) = (-s^2 + 5s + 17) / [(s + 6)(s - 1)^2]

Partial Fraction Decomposition

Now comes the fun part – partial fraction decomposition! This technique allows us to break down a complex rational expression into simpler fractions, making it easier to find the inverse Laplace transform. Think of it as taking a complicated dish and breaking it down into its individual ingredients.

  1. Set up the decomposition: We express Y(s) as a sum of simpler fractions:

    Y(s) = A/(s + 6) + B/(s - 1) + C/(s - 1)^2

    where A, B, and C are constants that we need to determine. It might look intimidating at first, but trust me, it's just a matter of following the steps.

  2. Clear the denominators: We multiply both sides of the equation by the common denominator extbf{((s + 6)(s - 1)^2)} to get rid of the fractions:

    -s^2 + 5s + 17 = A(s - 1)^2 + B(s + 6)(s - 1) + C(s + 6)

  3. Solve for the constants: There are a couple of ways to solve for A, B, and C. One common method is to substitute specific values of s that make some of the terms zero. Another method is to expand the right side, equate coefficients of like powers of s, and solve the resulting system of equations. Let's use a combination of both!

    • Let s = 1: This eliminates the A and B terms:

      -1 + 5 + 17 = C(1 + 6)

      21 = 7C

      C = 3

    • Let s = -6: This eliminates the B and C terms:

      -36 - 30 + 17 = A(-6 - 1)^2

      -49 = 49A

      A = -1

    • Equate coefficients of s^2: We look at the coefficients of the s^2 terms on both sides:

      -1 = A + B

      Since we know A = -1:

      -1 = -1 + B

      B = 0

  4. Substitute the values: Now that we've found A, B, and C, we plug them back into our partial fraction decomposition:

    Y(s) = -1/(s + 6) + 0/(s - 1) + 3/(s - 1)^2

    Y(s) = -1/(s + 6) + 3/(s - 1)^2

Inverse Laplace Transform

The final step is to take the inverse Laplace transform of Y(s) to get our solution y(t) back in the time domain. This is like reversing the initial transformation, going from the frequency domain back to the time domain.

  1. Apply inverse transform: We apply the inverse Laplace transform to each term. Remember these inverse Laplace transforms:

    • L^{-1}{1/(s - a)} = e^{at}
    • L^{-1}{1/(s - a)^2} = te^{at}

  2. Transform back to time domain: Applying these to our Y(s), we get:

    y(t) = L^{-1}{-1/(s + 6)} + L^{-1}{3/(s - 1)^2}

    y(t) = -e^{-6t} + 3te^t

Final Solution

And there you have it! We've successfully solved the differential equation using the Laplace transform. Our solution is:

y(t) = -e^{-6t} + 3te^t

This function satisfies both the original differential equation and the given initial conditions. It's a beautiful example of how the Laplace transform can turn a tricky problem into a manageable one. Remember, solving differential equations is like piecing together a puzzle – each step brings you closer to the final picture. Keep practicing, and you'll become a master of this powerful technique!

Conclusion

Solving differential equations with Laplace transforms might seem daunting at first, but with practice, it becomes a powerful tool in your mathematical arsenal. Remember, the key is to break down the problem into smaller, manageable steps. Transform the equation, solve for Y(s), use partial fraction decomposition, and then take the inverse Laplace transform. With a bit of patience and perseverance, you'll be solving differential equations like a pro in no time! Keep up the great work, and don't hesitate to tackle new challenges – you've got this!