Chapter 10: Problem 11
Solve the initial value problem \(y^{\prime \prime}+4 y=0, y(0)=\sqrt{3}, y^{\prime}(0)=2\).
Short Answer
Expert verified
The solution is \(y(t) = \sqrt{3} \cos(2t) + \sin(2t)\).
Step by step solution
01
Identifying the Differential Equation Type
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients, expressed as \(y'' + 4y = 0\).
02
Solving the Characteristic Equation
For the equation \(y'' + 4y = 0\), we assume a solution of the form \(y = e^{rt}\) and substitute this into the equation to get the characteristic equation \(r^2 + 4 = 0\).
03
Finding the Roots of the Characteristic Equation
Solve the characteristic equation \(r^2 + 4 = 0\). This gives \(r^2 = -4\), leading to the roots \(r = \pm 2i\).
04
Writing the General Solution
With the roots \(r = \pm 2i\), the general solution to the differential equation is \(y(t) = C_1 \cos(2t) + C_2 \sin(2t)\).
05
Applying Initial Conditions to Find Constants
Use the initial conditions to find \(C_1\) and \(C_2\). First, set \(t = 0\): \(y(0) = C_1\). Given \(y(0) = \sqrt{3}\), we have \(C_1 = \sqrt{3}\).
06
Differentiating the General Solution
Find \(y'(t)\) by differentiating the general solution: \(y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t)\).
07
Using the Second Initial Condition
Use the condition \(y'(0) = 2\) to find \(C_2\). At \(t = 0\), \(y'(0) = 2C_2 = 2\), which implies \(C_2 = 1\).
08
Writing the Particular Solution
Substitute \(C_1\) and \(C_2\) back into the general solution to get \(y(t) = \sqrt{3} \cos(2t) + \sin(2t)\). This is the particular solution to the initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-order Linear Homogeneous
A second-order linear homogeneous differential equation is a type of differential equation crucial in mathematical modeling, particularly in physics and engineering.
Such equations involve a second derivative of a function and can be written in the standard form:
\[ a y'' + b y' + c y = 0 \]where \(a\), \(b\), and \(c\) are constants. Important features of these equations include their linearity and lack of a non-zero term on the right-hand side.
This gives them the 'homogeneous' distinction, meaning the solution structure remains consistent, allowing the superposition of solutions. In this particular exercise, the equation is a simplified version: \(y'' + 4y = 0\), indicating constant coefficients with no first derivative term.
Such equations involve a second derivative of a function and can be written in the standard form:
\[ a y'' + b y' + c y = 0 \]where \(a\), \(b\), and \(c\) are constants. Important features of these equations include their linearity and lack of a non-zero term on the right-hand side.
This gives them the 'homogeneous' distinction, meaning the solution structure remains consistent, allowing the superposition of solutions. In this particular exercise, the equation is a simplified version: \(y'' + 4y = 0\), indicating constant coefficients with no first derivative term.
Characteristic Equation
The characteristic equation is a fundamental tool for solving second-order linear homogeneous differential equations with constant coefficients.
Assuming a solution in the exponential form \(y = e^{rt}\), substitute this into the homogeneous equation to derive the characteristic equation:
Assuming a solution in the exponential form \(y = e^{rt}\), substitute this into the homogeneous equation to derive the characteristic equation:
- \( y'' + 4y = 0 \)
- Substitute \(y = e^{rt}\): \(r^2 e^{rt} + 4e^{rt} = 0\)
- Thus, the characteristic equation becomes \( r^2 + 4 = 0 \)
Initial Value Problem
An initial value problem in differential equations refers to solving the equation with specific initial conditions provided.
Here, the initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) are used to find specific solutions for the constants in the general solution.
Here, the initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) are used to find specific solutions for the constants in the general solution.
- Step 1: Express the general solution as \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \)
- Step 2: Apply \( y(0) = \sqrt{3} \) to find \( C_1 = \sqrt{3} \)
- Step 3: Differentiate the general solution to find \( y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t) \)
- Step 4: Use \( y'(0) = 2 \) to establish \( C_2 = 1 \)
General Solution
The general solution of a differential equation represents a family of functions that includes all possible solutions to the equation.
Given the roots \( r = \pm 2i \), the general solution for the differential equation \( y'' + 4y = 0 \) is derived using complex exponentials:
Given the roots \( r = \pm 2i \), the general solution for the differential equation \( y'' + 4y = 0 \) is derived using complex exponentials:
- The solution takes the form \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \)
- This involves arbitrary constants \( C_1 \) and \( C_2 \), which can adapt to various initial conditions
Particular Solution
The particular solution is a single specific solution derived from the general solution by applying initial conditions.
From our problem, after identifying the general solution \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \), initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) were used to determine \( C_1 \) and \( C_2 \) as:
\[ y(t) = \sqrt{3} \cos(2t) + \sin(2t) \]This solution satisfies both the differential equation and the specified initial conditions, creating a singular path within the family of general solutions.
From our problem, after identifying the general solution \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \), initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) were used to determine \( C_1 \) and \( C_2 \) as:
- \( C_1 = \sqrt{3} \)
- \( C_2 = 1 \)
\[ y(t) = \sqrt{3} \cos(2t) + \sin(2t) \]This solution satisfies both the differential equation and the specified initial conditions, creating a singular path within the family of general solutions.