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Magazine Chapter 4: Problem 34
\(4 y^{\prime \prime}+y=0, \quad y(1)=0, \quad y^{\prime}(1)=-2\)
Short Answer
Expert verified
The solution is \(y(t) = -2 \cot \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}t\right) - 2 \csc \left(\frac{1}{2}\right) \sin \left(\frac{1}{2}t\right)\).
Step by step solution
01
Characteristic Equation
The given differential equation is \(4y'' + y = 0\). To solve this, we start by assuming a solution of the form \(y = e^{rt}\). Substituting this in, we get the characteristic equation: \(4r^2 + 1 = 0\).
02
Solve Characteristic Equation
Solve the characteristic equation \(4r^2 + 1 = 0\) to find \(r^2 = -\frac{1}{4}\). This gives \(r = \pm \frac{i}{2}\), which are complex roots.
03
Write General Solution
For complex roots \(r = \pm \frac{i}{2}\), the general solution is \(y(t) = c_1 \cos \left(\frac{1}{2}t\right) + c_2 \sin \left(\frac{1}{2}t\right)\).
04
Apply Initial Conditions
Use the initial conditions to find \(c_1\) and \(c_2\). Start with \(y(1) = 0\), which gives \(c_1 \cos \left(\frac{1}{2}\right) + c_2 \sin \left(\frac{1}{2}\right) = 0\).
05
Derivative and Second Condition
Find \(y'(t) = -\frac{1}{2}c_1 \sin \left(\frac{1}{2}t\right) + \frac{1}{2}c_2 \cos \left(\frac{1}{2}t\right)\). Use \(y'(1) = -2\) to setup \(-\frac{1}{2}c_1 \sin \left(\frac{1}{2}\right) + \frac{1}{2}c_2 \cos \left(\frac{1}{2}\right) = -2\).
06
Solve for Constants
Solve the two equations from Steps 4 and 5 simultaneously: \(c_1 \cos \left(\frac{1}{2}\right) + c_2 \sin \left(\frac{1}{2}\right) = 0\) and \(-\frac{1}{2}c_1 \sin \left(\frac{1}{2}\right) + \frac{1}{2}c_2 \cos \left(\frac{1}{2}\right) = -2\). Solve to find \(c_1 = -2\cot\left(\frac{1}{2}\right)\) and \(c_2 = -2\csc\left(\frac{1}{2}\right)\).
07
Final Solution
Substitute \(c_1\) and \(c_2\) back into the general solution to get:\[y(t) = -2 \cot \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}t\right) - 2 \csc \left(\frac{1}{2}\right) \sin \left(\frac{1}{2}t\right)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
When solving a homogeneous second-order differential equation like the given one, the first step is to find the characteristic equation. This equation is typically derived from assuming a solution of the form \( y = e^{rt} \). For the differential equation \( 4y'' + y = 0 \), we substitute \( y = e^{rt} \) and its derivatives into the equation.
- The derivative \( y'' = r^2 e^{rt} \) used in \( 4y'' + y = 0 \) gives the characteristic equation \( 4r^2 + 1 = 0 \).
- This characteristic equation is crucial as it simplifies the problem and helps us determine the nature of the roots \( r \).
Complex Roots
Upon solving the characteristic equation \( 4r^2 + 1 = 0 \), we find the roots to be \( r = \pm \frac{i}{2} \). These roots are complex. Complex roots typically appear in conjugate pairs, which is the case here.
- Complex roots lead to solutions involving trigonometric functions, such as sine and cosine.
- For a pair of complex roots \( \alpha \pm \beta i \), the solution to the differential equation is of the form: \( y(t) = c_1 e^{\alpha t} \cos(\beta t) + c_2 e^{\alpha t} \sin(\beta t) \).
Initial Conditions
Initial conditions are essential for finding the specific solution that satisfies our equation. They turn the general solution, which includes arbitrary constants, into a particular solution.
- For the equation \( y = c_1 \cos\left(\frac{1}{2}t\right) + c_2 \sin\left(\frac{1}{2}t\right) \), initial conditions are \( y(1) = 0 \) and \( y'(1) = -2 \).
- The condition \( y(1) = 0 \) leads to the equation \( c_1 \cos\left(\frac{1}{2}\right) + c_2 \sin\left(\frac{1}{2}\right) = 0 \).
- These conditions allow us to set up a system of equations, necessary for solving the unknown coefficients \( c_1 \) and \( c_2 \). Here it's important to note how the trigonometric terms help capture the natural occurrence of oscillatory behavior in solutions.
General Solution
A general solution to a differential equation with complex roots contains trigonometric functions because of Euler's formula. In this scenario, the general solution \( y(t) = c_1 \cos\left(\frac{1}{2}t\right) + c_2 \sin\left(\frac{1}{2}t\right) \) forms the basis of all possible solutions.
- The constants \( c_1 \) and \( c_2 \) are determined by applying initial conditions.
- Once these constants are found, substituting them back into the general form provides the particular solution.
