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Magazine Chapter 7: Problem 47
Obtain, from the particular solution indicated, the value of \(y\) and the value of \(y^{\prime}\) at \(x=2\). $$ 4 y^{\prime \prime}+y=2 ; \text { at } x=\pi, y=0, y^{\prime}=1. $$
Short Answer
Expert verified
At \(x=2\), \(y = 2 - 2 \sin(1)\) and \(y' = -\cos(1)\).
Step by step solution
01
Identify the Given Equation and Initial Conditions
We are given the differential equation \(4y'' + y = 2\), and initial conditions \(y(\pi) = 0\) and \(y'(\pi) = 1\). We need to find \(y\) and \(y'\) at \(x=2\).
02
Consider a General Solution
One approach to solving this differential equation is to assume a solution of the form \(y(x) = y_p(x) + y_h(x)\), where \(y_p(x)\) is a particular solution and \(y_h(x)\) is the solution to the homogeneous equation \(4y'' + y = 0\).
03
Solve the Homogeneous Equation
To solve the homogeneous differential equation \(4y'' + y = 0\), assume a solution of the form \(y_h = e^{rx}\). Substituting \(y_h\) into the homogeneous equation results in the characteristic equation \(4r^2 + 1 = 0\). Solving gives roots \(r = \pm \frac{i}{2}\), resulting in the complementary solution \(y_h = C_1 \cos\left(\frac{1}{2} x\right) + C_2 \sin\left(\frac{1}{2} x\right)\).
04
Find the Particular Solution
To find a particular solution \(y_p\), assume it is a constant \(y_p = A\) such that \(4(0) + A = 2\). This gives \(A = 2\), so the particular solution is \(y_p = 2\).
05
Combine Solutions
The general solution is \(y = y_h + y_p = C_1 \cos\left(\frac{1}{2} x\right) + C_2 \sin\left(\frac{1}{2} x\right) + 2\).
06
Apply Initial Conditions
Apply the initial conditions to find \(C_1\) and \(C_2\). Start with \(y(\pi) = 0\), leading to the equation \(C_1 \cos\left(\frac{1}{2} \pi\right) + C_2 \sin\left(\frac{1}{2} \pi\right) + 2 = 0\), giving \(C_2 + 2 = 0\) or \(C_2 = -2\).
07
Apply the First Derivative Initial Condition
Differentiate the general solution: \(y' = -\frac{1}{2}C_1 \sin\left(\frac{1}{2} x\right) + \frac{1}{2}C_2 \cos\left(\frac{1}{2} x\right)\). Apply the condition \(y'(\pi) = 1\) to get \(-\frac{1}{2}C_1 \cos\left(\frac{1}{2}\pi \right) - \frac{1}{2} \cdot 2 = 1\). Simplify to find \(C_1 = 0\).
08
Substitute and Evaluate at \(x=2\)
With \(C_1 = 0\) and \(C_2 = -2\), substitute into the solution: \(y(2) = 0 \cdot \cos\left(\frac{1}{2} \times 2\right) - 2 \cdot \sin\left(\frac{1}{2} \times 2\right) + 2\). Evaluate to find \(y(2) = -2 \sin(1) + 2\). Therefore, \(y(2) = 2 - 2 \sin(1)\).
09
Evaluate the Derivative at \(x=2\)
Now, evaluate the derivative \(y'(2) = -\frac{1}{2} \cdot 0 \cdot \cos(1) + \frac{1}{2} \cdot (-2) \cdot \cos(1) = -\cos(1)\). Therefore, \(y'(2) = -\cos(1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Particular Solution
When dealing with differential equations, finding a particular solution is crucial as it represents a specific solution that satisfies the non-homogeneous part of the equation. In the given problem, we need to solve the equation \(4y'' + y = 2\). This equation is non-homogeneous because of the constant term on the right side, which is \(2\). A simple approach to find a particular solution in such cases is to assume that the solution is a constant, say \(y_p = A\).
A constant particular solution is often chosen when the non-homogeneous part is itself a constant, allowing us to substitute \(y_p = A\) back into the equation to find that \(A = 2\).
A constant particular solution is often chosen when the non-homogeneous part is itself a constant, allowing us to substitute \(y_p = A\) back into the equation to find that \(A = 2\).
- The assumption makes the differential terms involve zero, simplifying the equation drastically.
- This results in the particular solution \(y_p = 2\) for this problem.
Homogeneous Equation
A homogeneous equation is a form of differential equation derived when the non-homogeneous part becomes zero. In this exercise, the homogeneous differential equation is obtained by writing \(4y'' + y = 0\). This equation represents the part of the differential equation that does not include the constant or any term dependent on \(x\).
To solve this, we assume that a solution of the form \(y_h = e^{rx}\) will satisfy the homogeneous equation. Hence, substitution leads us to the characteristic equation, which is \(4r^2 + 1 = 0\).
To solve this, we assume that a solution of the form \(y_h = e^{rx}\) will satisfy the homogeneous equation. Hence, substitution leads us to the characteristic equation, which is \(4r^2 + 1 = 0\).
- This yields complex roots \(r = \pm \frac{i}{2}\).
- The solution then involves trigonometric functions resulting in: \(y_h = C_1 \cos\left(\frac{1}{2} x \right) + C_2 \sin\left(\frac{1}{2} x \right)\).
Initial Conditions
Initial conditions are crucial for finding the particular values of the constants in the general solution of a differential equation. In this example, the initial conditions given are \(y(\pi) = 0\) and \(y'(\pi) = 1\). These values tell us the state of the solution and its derivative at a particular point, specifically \(x = \pi\).
We substitute these into the general solution to solve for the constants \(C_1\) and \(C_2\) that appear in the homogeneous part of our solution.
We substitute these into the general solution to solve for the constants \(C_1\) and \(C_2\) that appear in the homogeneous part of our solution.
- For \(y(\pi) = 0\), we derive an equation that helps us discover that \(C_2 = -2\).
- Similarly, using \(y'(\pi) = 1\), we find \(C_1 = 0\).
Characteristic Equation
The characteristic equation is a fundamental tool used for solving linear homogeneous differential equations with constant coefficients. It is derived from the assumption that the solution can be an exponential function \(y_h = e^{rx}\).
For our given differential equation \(4y'' + y = 0\), substituting \(e^{rx}\) into it results in \(4r^2 y_h + y_h = 0\), simplifying to the characteristic equation: \(4r^2 + 1 = 0\).
For our given differential equation \(4y'' + y = 0\), substituting \(e^{rx}\) into it results in \(4r^2 y_h + y_h = 0\), simplifying to the characteristic equation: \(4r^2 + 1 = 0\).
- Solving this characteristic equation involves solving \(4r^2 = -1\), revealing complex roots \(r = \pm \frac{i}{2}\).
- These roots lead us to the complementary solution consisting of cosine and sine functions.
