Chapter 15: Problem 9

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}+16 y=8 \cos 4 t, \quad y_{0}=0, \quad y_{0}^{\prime}=8 $$

Short Answer

Expert verified

The solution is $y(t) = 8 \cos(4t) + 16t \sin(4t)$

Step by step solution

Apply the Laplace Transform

Apply the Laplace transform to both sides of the differential equation. Recall that the Laplace transform of a second derivative is given by $ L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) $. For the given problem, $y(0) = 0$ and $y'(0) = 8$.

Substitute Initial Conditions

Substitute $y(0) = 0$ and $y'(0) = 8$ into the Laplace transform equation to get: \[ L\{y''\} + 16L\{y\} = 8L\{\cos 4t\} \Rightarrow (s^2Y(s) - 0 - 8) + 16Y(s) = 8 \frac{s}{s^2 + 16} \]

Solve for Y(s)

Simplify the equation to solve for $Y(s)$: \[ s^2Y(s) + 16Y(s) - 8 = 8 \frac{s}{s^2 + 16} \Rightarrow (s^2 + 16)Y(s) - 8 = 8 \frac{s}{s^2 + 16} \Rightarrow Y(s) = \frac{8 \frac{s}{s^2 + 16} + 8}{s^2 + 16} \Rightarrow Y(s) = \frac{8 \left( \frac{s}{s^2 + 16} + 1 \right)}{s^2 + 16} \]

Simplify the Expression

Simplify the expression further: \[ Y(s) = \frac{8 \left( \frac{s + s^2 + 16}{s^2 + 16} \right)}{s^2 + 16} \Rightarrow Y(s) = \frac{8}{s^2 + 16} \left(\frac{s + s^2 + 16}{s^2 + 16} \right) \Rightarrow Y(s) = \frac{8(s + s^2 + 16)}{(s^2 + 16)^2} \]

Take the Inverse Laplace Transform

Now, take the inverse Laplace transform to find $y(t)$: \[ L^{-1}\{Y(s)\} = L^{-1}\left\{ \frac{8s}{(s^2 + 16)^2} + \frac{128}{(s^2 + 16)^2} \right\} \]

Apply Known Inverse Transforms

Use the known inverse Laplace transforms: $L^{-1}\left( \frac{s}{s^2 + 16} \right) = \cos(4t) $ and $L^{-1}\left( \frac{1}{(s^2 + 16)^2} \right) = \frac{t}{8} \sin(4t) $ to get: \[ y(t) = 8 \cos(4t) + 16t \sin(4t) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

Differential equations are equations that relate a function with its derivatives. In this exercise, we dealt with a second-order linear differential equation: \ \[y'' + 16y = 8 \cos(4t)\] It means the function $y(t)$ and its second derivative $y''$ are related. These equations are essential in modeling real-world systems where the rate of change is proportional to other variables.

Initial Conditions

Initial conditions are values specified for the function and its derivatives at a starting point. In our problem, we had: \- $y(0) = 0$ \- $y'(0) = 8$ \These conditions help in solving the differential equation uniquely. Without them, there could be multiple solutions.

Inverse Laplace Transform

The inverse Laplace transform is used to find the original function from its Laplace transform. In this exercise, after manipulating the given differential equation and transforming it into the Laplace domain, we ended up with: \[L^{-1}\bigg\{ \frac{8s}{(s^2 + 16)^2} + \frac{128}{(s^2 + 16)^2} \bigg\}\] Applying known inverse Laplace transformations helped us get back to the time domain solution.

Cosine Function

The cosine function, $\cos(4t)$, appears in our differential equation. It is known that: \- $\cos(4t)$ is a periodic function oscillating with a frequency of $4$ and amplitude $1$. \- When transformed using Laplace, we get $L\{\cos(4t)\} = \frac{s}{s^2 + 16}$. \ We used this property to convert the differential equation to its Laplace form and later applied the inverse transform to find the solution.

Sine Function

The sine function, $sin(4t)$, stems from the inverse Laplace transform. After calculating the inverse of each term, we applied known transforms: \ $L^{-1}\bigg\{ \frac{1}{(s^2 + 16)^2} \bigg\} = \frac{t}{8} sin(4t)$. \ The sine function represents periodic oscillation, similar to the cosine but shifted by $\frac{\pi}{2}$ radians.

Second Derivative

The second derivative indicates the rate of change of the rate of change, showing how the function's slope is changing. In Laplace terms, the second derivative is: \ \[L\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)\] \In our problem, with the initial conditions $y(0) = 0, y'(0) = 8$, it becomes \[L\{y''\} = s^2 Y(s) - 8\]. This was part of transforming the original differential equation before solving for $Y(s)$.