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By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+9 y=\cos 3 t, \quad y_{0}=2, y_{0}^{\prime}=0$$

Short Answer

Expert verified
Using Laplace transforms, the solution is: \( y(t)=2 \cos(3t) + \frac{19}{9} \sin(3t) \)

Step by step solution

01

Apply the Laplace Transform to the Differential Equation

Taking the Laplace transform of both sides of the equation y'' + 9y = \cos(3t)\,we get \(\mathcal{L}\{y'' + 9y\} = \mathcal{L}\{\cos(3t)\}\). Utilizing the linearity of the Laplace transform, this becomes \(\mathcal{L}\{y''\} + 9\mathcal{L}\{y\} = \mathcal{L}\{\cos(3t)\}\).
02

Use Laplace Transform Properties

Recall that \(\mathcal{L}\{y''\} = s^2\mathcal{L}\{y\} - sy(0) - y'(0)\). Using the initial conditions \(y(0)=2\) and \(y'(0)=0\), we substitute to obtain \(s^2Y(s) - 2s + 0 + 9Y(s) = \mathcal{L}\{\cos(3t)\}\).
03

Laplace Transform of \(\cos(3t)\)

Recall that \(\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2+9}\). Thus, the equation becomes\(s^2Y(s) - 2s + 9Y(s) = \frac{s}{s^2 + 9}\).
04

Solve for \(Y(s)\)

Combine like terms:\[(s^2 + 9) Y(s) - 2s = \frac{s}{s^2 + 9}\]. Factor to isolate \(Y(s)\):\[(s^2 + 9)Y(s) = \frac{s}{s^2 + 9} + 2s\]. Thus,\[Y(s) = \frac{s/(s^2 + 9) + 2s}{s^2 + 9}\].
05

Simplify \(Y(s)\)

Simplify the right-hand side:\[Y(s) = \frac{s + 2s(s^2 + 9)}{(s^2 + 9)^2} = \frac{s + 2s^3 + 18s}{(s^2 + 9)^2}\].Combine like terms in the numerator:\[Y(s) = \frac{2s^3 + 19s}{(s^2 + 9)^2}\].
06

Use the Inverse Laplace Transform

Split into partial fractions and find the inverse Laplace transform:\[Y(s) = \frac{2s^3}{(s^2 + 9)^2} + \frac{19s}{(s^2 + 9)^2}\].Using the inverse Laplace transform properties, \[\mathcal{L}^{-1}\left\{ \frac{2s^3}{(s^2 + 9)^2} \right \} = 2 \cos(3t), \mathcal{L}^{-1}\left\{ \frac{19s}{(s^2 + 9)^2} \right \} = \frac{19}{9} \sin(3t)\].Thus,\[y(t) = 2 \cos(3t) + \frac{19}{9} \sin(3t)\].

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Key Concepts

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

Differential Equations
Differential equations involve equations that include derivatives of a function. They are essential for modeling various real-world phenomena, from physics to biology. In the context of the provided exercise, we are dealing with a second-order linear differential equation given by:

\( y'' + 9y = \cos(3t) \).

The goal is to find the function \( y(t) \) that satisfies the differential equation and the initial conditions.
Initial Conditions
Initial conditions are given values of the function and its derivatives at a specific point, often at \( t = 0 \). They help uniquely determine the solution to a differential equation. For example, in our exercise:
  • \( y(0) = 2 \)
  • \( y'(0) = 0 \)
These conditions are used in the Laplace transform steps to simplify and solve for \( Y(s) \).
Inverse Laplace Transform
The inverse Laplace transform is used to convert a function in the Laplace domain back to the time domain. After solving the transformed differential equation for \( Y(s) \), we need to find the original function \( y(t) \). The Laplace transform of a function \( f(t) \) is typically denoted as \( \mathcal{L} \{ f(t) \} \), and its inverse is denoted as \( \mathcal{L}^{-1} \{ F(s) \} \).

In our solution, we use inverse Laplace transforms to convert terms like
  • \( \mathcal{L}^{-1} \left\{ \frac{2s^3}{(s^2 + 9)^2} \right\ \} \right = 2 \cos (3t) \)
  • \( \mathcal{L}^{-1} \left\{ \frac{19s}{(s^2 + 9)^2} \right\ \} = \frac{19}{9} \sin(3t) \)
resulting in \( y(t) = 2 \cos(3t) +\ \frac{19}{9} \sin(3t) \).
Partial Fractions
Partial fractions are used to break down complex rational expressions into simpler fractions, making them easier to invert using inverse Laplace transforms. In our problem, to solve for \( Y(s) \), we express it as:
  • \( Y(s) = \frac{2s^3}{(s^2 + 9)^2} + \frac{19s}{(s^2 + 9)^2} \)
This decomposition allows us to handle each term individually when finding the inverse Laplace transform.
Trigonometric Functions
Trigonometric functions often appear in differential equations and their solutions, particularly in problems involving oscillatory behavior. Common functions include sine and cosine. In our solution:
  • \( \cos (3t) \)
  • \( \sin (3t) \)
These functions are pivotal in forming the final solution \( y(t) \). Trigonometric identities and their properties are used effectively when dealing with Laplace and inverse Laplace transforms.

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Most popular questions from this chapter

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

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\left(x \cos y-e^{-\sin y}\right) d y+d x=0$$

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \(y^{\prime}+2 x y^{2}=0, \quad y=1\) when \(x=2\)

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is \(2 \pi \sqrt{h / g}\) where \(h\) is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.

(a) A rocket of (variable) mass \(m\) is propelled by steadily ejecting part of its mass at velocity \(u\) (constant with respect to the rocket). Neglecting gravity, the differential equation of the rocket is \(m(d v / d m)=-u\) as long as \(v \ll c, c=\) speed of light. Find \(v\) as a function of \(m\) if \(m=m_{0}\) when \(v=0\). (b) In the relativistic region ( \(v / c\) not negligible), the rocket equation is \(m \frac{d v}{d m}=-u\left(1-\frac{v^{2}}{c^{2}}\right)\). Solve this differential equation to find \(v\) as a function of \(m .\) Show that \(v / c=\) \((1-x) /(1+x),\) where \(x=\left(m / m_{0}\right)^{2 u / c}\).

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