Chapter 8: Problem 4
By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+y=\sin t, \quad y_{0}=1, y_{0}^{\prime}=0$$
Short Answer
Expert verified
y(t) = \frac{t \sin t}{2} + \cos t
Step by step solution
01
Take the Laplace transform of both sides of the differential equation
Apply the Laplace transform to the given differential equation \[ y'' + y = \sin t \] Using the linearity property of Laplace transforms, we get \[ \mathcal{L}\{y''\} + \mathcal{L}\{y\} = \mathcal{L}\{\sin t\} \]
02
Use the Laplace transforms of the derivatives
Recall that the Laplace transform of the first derivative is \[ \mathcal{L}\{y'\} = sY(s) - y(0) \] and for the second derivative, it is \[ \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) \] Here, initial conditions are given as \( y(0) = 1 \) and \( y'(0) = 0 \). Substitute these into the Laplace transforms.
03
Substitute the transforms and initial conditions
Substituting the Laplace transforms and initial conditions into the differential equation, we get: \[ s^2Y(s) - s \cdot 1 - 0 + Y(s) = \mathcal{L}\{\sin t\} \] Using \( \mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1} \), the equation simplifies to: \[ (s^2 + 1)Y(s) - s = \frac{1}{s^2 + 1} \]
04
Solve for Y(s)
Isolate \( Y(s) \): \[ (s^2 + 1)Y(s) = \frac{1}{s^2 + 1} + s \] \[ Y(s) = \frac{1}{(s^2 + 1)^2} + \frac{s}{s^2 + 1} \]
05
Find inverse Laplace transform
Find the inverse Laplace transform to get \( y(t) \): The inverse Laplace transform of \( \frac{1}{(s^2 + 1)^2} \) is \( \frac{t \sin t}{2} \) and \( \frac{s}{s^2 + 1} \) is \( \cos t \). Thus, \[ y(t) = \frac{t \sin t}{2} + \cos t \]
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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 involve derivatives of a function. They describe how a certain quantity changes over time. For instance, if you have a differential equation like \[ y'' + y = \sin t, \], it explains how the second derivative of \(y\), noted as \(y''\), and \(y\) itself relate to the sine of \(t\). These equations are vital in modeling various real-world phenomena ranging from physics to biology.
Initial Conditions
Initial conditions are the values of the function and its derivatives at the start (usually when t=0). They are essential for solving differential equations because they help determine the specific solution out of the many possible ones. For example, in the differential equation we considered, the initial conditions are \(y(0) = 1\) and \(y'(0) = 0\). These tell us the state of the system at the beginning and are used to find the constants needed in the solution.
Inverse Laplace Transform
The inverse Laplace transform is the process of converting a Laplace-transformed function back into its original time-domain form. The Laplace transform makes solving differential equations easier because it converts them into algebraic equations. To get back to the time domain, we use the inverse Laplace transform. In our example, after finding \(Y(s)\), the inverse Laplace transform helped us reach the final solution for \(y(t)\), which is \(y(t) = \frac{t \sin t}{2} + \cos t\).
Linear Differential Equations
A linear differential equation is one in which the function and its derivatives appear to the power of one and are not multiplied together. Our equation \(y'' + y = \sin t\) is linear because it fits this form. Linear differential equations are easier to solve and have many tools dedicated to them, such as the Laplace transform. They appear often in engineering and the sciences because they are good at modeling many physical systems.