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Mobile App Chapter 15: Problem 4
By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}+y=\sin t, \quad y_{0}=1, \quad y_{0}^{\prime}=0 $$
Short Answer
Expert verified
The solution is \(y(t) = \cos t + \frac{1}{2} t \sin t \).
Step by step solution
01
Apply Laplace Transform to Both Sides
Take the Laplace transform of the entire differential equation \(y'' + y = \sin t\) to get: \( \mathcal{L} \{ y'' + y \} = \mathcal{L} \{ \sin t \} \).
02
Use Laplace Transform Properties
Using the properties of the Laplace transform, we have: \(\mathcal{L} \{ y'' \} = s^2 Y(s) - sy(0) - y'(0)\) and \(\mathcal{L} \{ y \} = Y(s)\).
03
Substitute Initial Conditions
Use the initial conditions \(y(0) = 1\) and \(y'(0) = 0\): \(\mathcal{L} \{ y'' \} = s^2 Y(s) - s \cdot 1 - 0 = s^2 Y(s) - s\).
04
Form the Algebraic Equation
Substitute the transformations back into the original equation: \(s^2 Y(s) - s + Y(s) = \mathcal{L} \{ \sin t \} \).
05
Laplace Transform of Sine
Recall that \(\mathcal{L} \{ \sin t \} = \frac{1}{s^2 + 1} \). Thus, the equation becomes: \(s^2 Y(s) - s + Y(s) = \frac{1}{s^2 + 1} \).
06
Solve for Y(s)
Combine like terms to solve for \(Y(s)\): \((s^2 + 1)Y(s) = s + \frac{1}{s^2 + 1} \), and \(Y(s) = \frac{s}{s^2 + 1} + \frac{1}{(s^2 + 1)^2} \).
07
Inverse Laplace Transform
Find the inverse Laplace transform \(y(t) = \mathcal{L}^{-1} \{ Y(s) \}\): \(Y(s) = \mathcal{L}^{-1} \{\frac{s}{s^2 + 1} \} + \mathcal{L}^{-1} \{ \frac{1}{(s^2 + 1)^2} \} \).
08
Evaluate the Inverse Laplace Transforms
Using standard Laplace inverse tables, we find: \(\mathcal{L}^{-1} \{ \frac{s}{s^2 + 1} \} = \cos t \) and \(\mathcal{L}^{-1} \{ \frac{1}{(s^2 + 1)^2} \} = \frac{1}{2}t \sin t \).
09
Combine Results
Therefore, the final solution is: \(y(t) = \cos t + \frac{1}{2} t \sin t \).
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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 mathematical equations that involve an unknown function and its derivatives. They are essential in modeling various physical, biological, and engineering systems. In this exercise, we are given a second-order differential equation: \(y'' + y = \sin t\), where \(y''\) represents the second derivative of y with respect to t. This type of equation is common in the study of oscillatory systems, such as springs or circuits.
To solve this differential equation, we applied the Laplace transform, which converts the differential equation into an algebraic equation in the Laplace domain. This simplifies the process of finding the solution by turning differentiation into multiplication, making it easier to solve.
To solve this differential equation, we applied the Laplace transform, which converts the differential equation into an algebraic equation in the Laplace domain. This simplifies the process of finding the solution by turning differentiation into multiplication, making it easier to solve.
Initial Conditions
Initial conditions are the values of the function and its derivatives at the beginning of the interval of interest. They are crucial for determining a unique solution to the differential equation. In this problem, the initial conditions are given as:
- \(y(0) = 1\) - the value of the function y at \(t = 0\)
- \(y'(0) = 0\) - the value of the first derivative of y at \(t = 0\)
Inverse Laplace Transform
The inverse Laplace transform is the process of converting an algebraic expression in the Laplace domain back to the time domain. After obtaining Y(s) in the Laplace domain, we need its inverse Laplace transform to determine the original time-domain function y(t).
- We used the inverse Laplace transform of \(\frac{s}{s^2 + 1}\) to get \(\cos t\)
- We used the inverse Laplace transform of \(\frac{1}{(s^2 + 1)^2}\) to get \(\frac{1}{2} t \sin t\)
Sine Function Transformation
The sine function transformation is an essential part of solving differential equations using the Laplace transform. In this exercise, we dealt with \(\sin t\), which needs to be transformed into the Laplace domain. The Laplace transform of \(\sin t\) is \(\frac{1}{s^2 + 1}\). This transformation allows us to include the effect of the sine function within the algebraic equation we solve.
Recognizing the Laplace transforms of common functions such as sine is integral for solving differential equations using this method. It simplifies expressions and enables the application of algebraic manipulation to find the solution.
By understanding these transformations, students can solve a wide variety of differential equations involving trigonometric functions.
Recognizing the Laplace transforms of common functions such as sine is integral for solving differential equations using this method. It simplifies expressions and enables the application of algebraic manipulation to find the solution.
By understanding these transformations, students can solve a wide variety of differential equations involving trigonometric functions.
