Problem 25 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given initial-value problem. \(y^{\prime \prime}+4 y=9 \sin t, \quad y(0)=1, \quad y^{\prime}(0)=-1\).

Step-by-Step Solution

Answer

The solution to the given initial-value problem is \(y(t) = \frac{3}{8}\cos(t) - \frac{3}{2}\sin(t) + \frac{5}{8}\cos(2t) + \frac{9}{2}\sin(2t)\).

Step 1: Laplace Transform of the given Initial-value problem

First, let's take the Laplace Transform of the given differential equation: \[y''(t) + 4y(t) = 9\sin(t)\] Applying Laplace Transform, we get: \[\mathcal{L}\{y''(t) + 4y(t)\} = \mathcal{L}\{9\sin(t)\}\]

Step 2: Laplace Transform of individual terms

Now, let's find the Laplace Transform of each term individually: Laplace Transform of \(y''(t)\) is \(s^2Y(s) - sy(0) - y'(0)\), and Laplace Transform of \(4y(t)\) is \(4Y(s)\). Also, Laplace Transform of \(9\sin(t)\) is \(\frac{9}{s^2+1}\). We already know that \(y(0) = 1\) and \(y'(0) = -1\). Now substituting these in the equation obtained in Step 1, we get \[(s^2Y(s) - s(1) - (-1)) + 4Y(s) = \frac{9}{s^2 + 1}\]

Step 3: Solve the equation for Y(s)

Simplify the equation and isolate Y(s): \[s^2Y(s) + 4Y(s) = \frac{9 + s^3 + s}{s^2 + 1}\] Now, divide both sides by \((s^2+1)\) and simplify: \[Y(s)(s^2+4) = \frac{s^3 + s + 9}{s^2 + 1}\] Now, divide both sides by \((s^2 + 4)\) to isolate Y(s): \[Y(s) = \frac{s^3 + s + 9}{(s^2 + 1)(s^2 + 4)}\]

Step 4: Inverse Laplace Transform

Now, to obtain \(y(t)\), we need to compute the inverse Laplace Transform of \(Y(s)\): \[y(t) = \mathcal{L}^{-1}\left\{\frac{s^3 + s + 9}{(s^2 + 1)(s^2 + 4)}\right\}\] To represent the rational function as a sum of partial fractions, we will rewrite it in the following form: \[Y(s) = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}\]

Step 5: Determine coefficients and simplify

Now we must find A, B, C, and D from the equation above and then compute the inverse Laplace Transform of the partial fractions. 1. Write the partial fraction decomposition as a single fraction: \[(As + B)(s^2 + 4) + (Cs + D)(s^2 + 1) = s^3 + s + 9\] 2. Expanding and collecting terms, we get \[[(A+C)s^3 + (A+C)s] + [(4A + B)s^2 + (4B + D)] = s^3 + s + 9\] 3. Compare the coefficients of each term to find A, B, C, and D: \(A+C = 1, 4A+B=0, A+C= 1, 4B+D=9\) Solving these equations, we get \(A = \frac{3}{8}, B = -\frac{3}{2}, C = \frac{5}{8}, D = 9\) 4. Now rewrite Y(s) using the found coefficients: \[Y(s) = \frac{\frac{3}{8}s - \frac{3}{2}}{s^2 + 1} + \frac{\frac{5}{8}s + 9}{s^2 + 4}\] 5. Compute the inverse Laplace Transform of these partial fractions \[y(t) = \mathcal{L}^{-1}\left\{\frac{\frac{3}{8}s - \frac{3}{2}}{s^2 + 1}\right\} + \mathcal{L}^{-1}\left\{\frac{\frac{5}{8}s + 9}{s^2 + 4}\right\}\] 6. Using the properties of inverse Laplace Transform, we get \[y(t) = \frac{3}{8}\cos(t) - \frac{3}{2}\sin(t) + \frac{5}{8}\cos(2t) + \frac{9}{2}\sin(2t)\] So, the solution to the given initial-value problem is: \[y(t) = \frac{3}{8}\cos(t) - \frac{3}{2}\sin(t) + \frac{5}{8}\cos(2t) + \frac{9}{2}\sin(2t)\]

Key Concepts

Initial-Value Problem

An initial-value problem involves differential equations where you need to find a function that satisfies certain conditions at the start (the initial values). In our exercise, we want to solve the differential equation \[y''(t) + 4y(t) = 9\sin(t)\]subject to the given initial conditions:

• \(y(0) = 1\) - This means at time \(t = 0\) the function \(y(t)\) equals 1.
• \(y'(0) = -1\) - This means at time \(t = 0\), the derivative of the function, or the rate of change, is -1.

These conditions help uniquely determine the solution for our differential equation. Without them, there might be infinite solutions. The initial values define how your solution behaves at the start, steering the pathway your function will take.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify complex rational expressions. In the context of solving differential equations with Laplace transforms, it helps in expressing a complicated function as a sum of simpler fractions. This makes it easier to apply inverse Laplace transforms.For the expression \[Y(s) = \frac{s^3 + s + 9}{(s^2 + 1)(s^2 + 4)}\], partial fraction decomposition involves rewriting it in the form:\[Y(s) = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}\]Here, \(A\), \(B\), \(C\), and \(D\) are constants that we find by setting up equations based on the identity:\[(As + B)(s^2 + 4) + (Cs + D)(s^2 + 1) = s^3 + s + 9\]This technique is particularly useful because it allows the inverse Laplace transformation of each simpler part separately, aiding in the step-by-step solution of the differential equation.

Inverse Laplace Transform

The inverse Laplace Transform is pivotal in finding the actual function \(y(t)\) from its transformed version \(Y(s)\). After decomposing \(Y(s)\) into partial fractions, you need to change these simpler expressions back into their respective time-domain versions.Applying the inverse Laplace transform, we compute:\[y(t) = \mathcal{L}^{-1}\left\{\frac{\frac{3}{8}s - \frac{3}{2}}{s^2 + 1}\right\} + \mathcal{L}^{-1}\left\{\frac{\frac{5}{8}s + 9}{s^2 + 4}\right\}\]Here, we use:

• The transform \(\mathcal{L}^{-1}\left\{\frac{a}{s^2 + 1}\right\} = a\cos(t)\) helps to revert cosine-related terms.
• For sine terms, \(\mathcal{L}^{-1}\left\{\frac{b}{s^2 + \omega^2}\right\} = b\sin(\omega t)\) assists in finding the solution.

In the exercise, following these principles helps translate the expression back into the desired time-domain function, resulting in the solution:\[y(t) = \frac{3}{8}\cos(t) - \frac{3}{2}\sin(t) + \frac{5}{8}\cos(2t) + \frac{9}{2}\sin(2t)\]