Solving Third-Order Initial Value Problems Using Laplace Transforms

This article will guide you through the process of solving a third-order initial value problem using the powerful method of Laplace transforms. Laplace transforms provide a systematic approach to solving linear differential equations, especially those with constant coefficients, by converting them into algebraic equations that are easier to manipulate. We will consider the following initial value problem:

$y^{\prime \prime \prime}+5 y^{\prime \prime}+2 y^{\prime}-8 y=-32, y(0)=6, y^{\prime}(0)=-13, y^{\prime \prime}(0)=77$

This problem involves a third-order linear homogeneous differential equation with constant coefficients, along with initial conditions for the function and its first two derivatives. The Laplace transform method will allow us to transform this differential equation into an algebraic equation in the Laplace domain, solve for the Laplace transform of the solution, and then apply the inverse Laplace transform to obtain the solution in the time domain.

1. Applying the Laplace Transform

To begin, we apply the Laplace transform to both sides of the differential equation. Recall that the Laplace transform of a function y(t), denoted by Y(s), is defined as:

$L{y(t)} = Y(s) = \int_{0}^{\infty} e^{-st}y(t) dt$

Using the properties of Laplace transforms, we can transform the derivatives of y(t) as follows:

• $L{y^{\prime}(t)} = sY(s) - y(0)$
• $L{y^{\prime \prime}(t)} = s^2Y(s) - sy(0) - y^{\prime}(0)$
• $L{y^{\prime \prime \prime}(t)} = s^3Y(s) - s^2y(0) - sy^{\prime}(0) - y^{\prime \prime}(0)$

Applying these properties to our differential equation, we get:

$L{y^{\prime \prime \prime} + 5y^{\prime \prime} + 2y^{\prime} - 8y} = L{-32}$

$L{y^{\prime \prime \prime}} + 5L{y^{\prime \prime}} + 2L{y^{\prime}} - 8L{y} = -32L{1}$

Substituting the Laplace transforms of the derivatives and the constant function, we obtain:

$[s^3Y(s) - s^2y(0) - sy^{\prime}(0) - y^{\prime \prime}(0)] + 5[s^2Y(s) - sy(0) - y^{\prime}(0)] + 2[sY(s) - y(0)] - 8Y(s) = -32(\frac{1}{s})$

Now, we substitute the given initial conditions y(0) = 6, y'(0) = -13, and y''(0) = 77:

$[s^3Y(s) - 6s^2 + 13s - 77] + 5[s^2Y(s) - 6s + 13] + 2[sY(s) - 6] - 8Y(s) = -\frac{32}{s}$

This step transforms the differential equation into an algebraic equation in the s-domain. The key is to correctly apply the Laplace transform properties and substitute the initial conditions. This algebraic equation will be solved for Y(s) in the next step. This process is a crucial step in simplifying the original differential equation into a more manageable form.

2. Solving for Y(s)

Now, let's simplify the equation and solve for Y(s). We start by expanding and collecting terms:

$s^3Y(s) - 6s^2 + 13s - 77 + 5s^2Y(s) - 30s + 65 + 2sY(s) - 12 - 8Y(s) = -\frac{32}{s}$

Combining the terms involving Y(s) and the constant terms, we get:

$(s^3 + 5s^2 + 2s - 8)Y(s) - 6s^2 - 17s - 24 = -\frac{32}{s}$

Next, we isolate the term with Y(s):

$(s^3 + 5s^2 + 2s - 8)Y(s) = 6s^2 + 17s + 24 - \frac{32}{s}$

To solve for Y(s), we divide both sides by the polynomial (s³ + 5s² + 2s - 8):

$Y(s) = \frac{6s^2 + 17s + 24 - \frac{32}{s}}{s^3 + 5s^2 + 2s - 8}$

To simplify further, we multiply the numerator and denominator by s to eliminate the fraction in the numerator:

$Y(s) = \frac{6s^3 + 17s^2 + 24s - 32}{s(s^3 + 5s^2 + 2s - 8)}$

Now, we need to factor the denominator. Notice that s³ + 5s² + 2s - 8 can be factored by observing that s = 1 is a root. Using synthetic division or polynomial long division, we find:

$s^3 + 5s^2 + 2s - 8 = (s - 1)(s^2 + 6s + 8)$

Further factoring the quadratic term, we get:

$s^2 + 6s + 8 = (s + 2)(s + 4)$

So, the denominator becomes:

$s(s - 1)(s + 2)(s + 4)$

Thus, our expression for Y(s) is:

$Y(s) = \frac{6s^3 + 17s^2 + 24s - 32}{s(s - 1)(s + 2)(s + 4)}$

This is the Laplace transform of the solution y(t). The next step is to perform partial fraction decomposition on Y(s) to prepare it for the inverse Laplace transform. This decomposition will break down the complex fraction into simpler fractions, each of which has a known inverse Laplace transform.

3. Partial Fraction Decomposition

To find the solution y(t), we need to apply the inverse Laplace transform to Y(s). To do this, we first perform partial fraction decomposition on the expression:

$Y(s) = \frac{6s^3 + 17s^2 + 24s - 32}{s(s - 1)(s + 2)(s + 4)}$

We decompose Y(s) into the form:

$Y(s) = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s + 2} + \frac{D}{s + 4}$

To find the constants A, B, C, and D, we multiply both sides by the denominator s(s - 1)(s + 2)(s + 4):

$6s^3 + 17s^2 + 24s - 32 = A(s - 1)(s + 2)(s + 4) + Bs(s + 2)(s + 4) + Cs(s - 1)(s + 4) + Ds(s - 1)(s + 2)$

Now, we solve for the constants by substituting convenient values of s:

• For s = 0:

  $-32 = A(-1)(2)(4) \Rightarrow -32 = -8A \Rightarrow A = 4$

• For s = 1:

  $6 + 17 + 24 - 32 = B(1)(3)(5) \Rightarrow 15 = 15B \Rightarrow B = 1$

• For s = -2:

  $6(-8) + 17(4) + 24(-2) - 32 = C(-2)(-3)(2) \Rightarrow -48 + 68 - 48 - 32 = 12C \Rightarrow -60 = 12C \Rightarrow C = -5$

• For s = -4:

  $6(-64) + 17(16) + 24(-4) - 32 = D(-4)(-5)(-2) \Rightarrow -384 + 272 - 96 - 32 = -40D \Rightarrow -240 = -40D \Rightarrow D = 6$

So, we have A = 4, B = 1, C = -5, and D = 6. Thus, the partial fraction decomposition is:

$Y(s) = \frac{4}{s} + \frac{1}{s - 1} - \frac{5}{s + 2} + \frac{6}{s + 4}$

This decomposition breaks Y(s) into simpler terms, each of which can be easily transformed back to the time domain using the inverse Laplace transform.

4. Applying the Inverse Laplace Transform

Now that we have decomposed Y(s) into simpler fractions, we can apply the inverse Laplace transform to each term. Recall the following basic inverse Laplace transforms:

• $L^{-1}{\frac{1}{s}} = 1$
• $L^{-1}{\frac{1}{s - a}} = e^{at}$

Using these, we can find the inverse Laplace transform of Y(s):

$y(t) = L^{-1}{Y(s)} = L^{-1}{\frac{4}{s} + \frac{1}{s - 1} - \frac{5}{s + 2} + \frac{6}{s + 4}}$

Applying the inverse Laplace transform to each term:

$y(t) = 4L^{-1}{\frac{1}{s}} + L^{-1}{\frac{1}{s - 1}} - 5L^{-1}{\frac{1}{s + 2}} + 6L^{-1}{\frac{1}{s + 4}}$

Using the known transforms, we get:

$y(t) = 4(1) + e^{t} - 5e^{-2t} + 6e^{-4t}$

Thus, the solution to the initial value problem is:

$y(t) = 4 + e^{t} - 5e^{-2t} + 6e^{-4t}$

This function y(t) satisfies the given differential equation and the initial conditions. The Laplace transform method has allowed us to solve a complex third-order differential equation by converting it into an algebraic problem in the s-domain, solving for Y(s), and then transforming back to the time domain using partial fraction decomposition and inverse Laplace transforms. This is a powerful technique for solving linear differential equations, especially those with constant coefficients and initial conditions.

Summary

In summary, we have successfully solved the third-order initial value problem using Laplace transforms. The key steps include:

1. Applying the Laplace transform to the differential equation.
2. Solving for Y(s) in the s-domain.
3. Performing partial fraction decomposition on Y(s).
4. Applying the inverse Laplace transform to obtain the solution y(t).

The final solution is:

$y(t) = 4 + e^{t} - 5e^{-2t} + 6e^{-4t}$

This method provides a systematic approach for solving linear differential equations with constant coefficients, and it is a valuable tool in engineering and physics.

For further exploration and practice, consider the following resources:

• Textbooks on differential equations and Laplace transforms.
• Online tutorials and video lectures on Laplace transforms.
• Practice problems and solutions available on various educational websites.

By mastering the technique of Laplace transforms, you can tackle a wide range of differential equations and related problems in various fields of science and engineering.