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Magazine Chapter 4: Problem 22
\(y^{\prime \prime}+36 y=0, y(0)=2, y^{\prime}(0)=-6\)
Short Answer
Expert verified
The solution is \( y(x) = 2 \cos(6x) - \sin(6x) \).
Step by step solution
01
- Identify the type of differential equation
The given equation is a second-order linear homogeneous differential equation: \( y'' + 36y = 0 \)
02
- Write the characteristic equation
For the differential equation \( y'' + 36y = 0 \), the characteristic equation is formed by replacing \(y''\) with \(r^2\) and \(y\) with 1: \( r^2 + 36 = 0 \)
03
- Solve the characteristic equation
Solve the characteristic equation \( r^2 + 36 = 0 \): \( r^2 = -36 \) This gives us: \( r = \pm 6i \)
04
- Write the general solution
The general solution for a differential equation with complex roots \( r = a \pm bi \) is: \( y(x) = e^{ax} (C_1 \cos(bx) + C_2 \sin(bx)) \).In this case, \(a = 0\) and \(b = 6\), so the general solution is: \( y(x) = C_1 \cos(6x) + C_2 \sin(6x) \)
05
- Apply the initial conditions
Initial condition \( y(0) = 2 \): \( y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 2 \). Hence, \( C_1 = 2 \).Initial condition \( y'(0) = -6 \): Differentiate \( y \) to get \( y' = -6C_1 \sin(6x) + 6C_2 \cos(6x) \).At \( x = 0 \): \( y'(0) = -6C_1 \sin(0) + 6C_2 \cos(0) = 6C_2 = -6 \). Hence, \( C_2 = -1 \).
06
- Write the specific solution
Substitute \( C_1 \) and \( C_2 \) back into the general solution: \( y(x) = 2 \cos(6x) - \sin(6x) \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
A characteristic equation is formed by converting a given differential equation into an algebraic equation. For the differential equation \( y'' + 36y = 0 \), replace \( y'' \) with \( r^2 \) and \( y \) with 1. This gives us the characteristic equation:
- \( r^2 + 36 = 0 \).
Complex Roots
In the characteristic equation, we had \( r^2 + 36 = 0 \). Solving this, we get:
- \( r^2 = -36 \) 鈫� \( r = \pm 6i \)
General Solution
When a differential equation has complex roots \(a \pm bi\), the general solution takes the form:
- \( y(x) = e^{ax} (C_1 \cos(bx) + C_2 \sin(bx)) \)
- \( y(x) = C_1 \cos(6x) + C_2 \sin(6x) \)
Initial Conditions
Initial conditions are additional constraints given to find specific values for the constants in the general solution. Here, we had:
1. \( y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 2 \). So, \( C_1 = 2 \).
2. Differentiate to get \( y' = -6C_1 \sin(6x) + 6C_2 \cos(6x) \). Then use \( y'(0) = -6 \):
- \( y(0) = 2 \)
- \( y'(0) = -6 \)
1. \( y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 2 \). So, \( C_1 = 2 \).
2. Differentiate to get \( y' = -6C_1 \sin(6x) + 6C_2 \cos(6x) \). Then use \( y'(0) = -6 \):
- \( 0 = 6C_2 - 6 \), so \( C_2 = -1 \).
- \( y(x) = 2 \cos(6x) - \sin(6x) \)
