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Magazine Chapter 8: Problem 39
Find the particular solution indicated. $$ \left(D^{2}+4 D+5\right) y=10 e^{-3 x} ; \text { when } x=0, y=4, y^{\prime}=0 $$
Short Answer
Expert verified
The particular solution is \(y = e^{-2x}(6 \cos x + 3 \sin x) - 2e^{-3x}\).
Step by step solution
01
Determine the Characteristic Equation
Identify the characteristic equation associated with the differential operator \(D^2 + 4D + 5\). The characteristic equation is obtained by setting \(D = m\) and solving \(m^2 + 4m + 5 = 0\).
02
Solve the Characteristic Equation
Solve the characteristic equation \(m^2 + 4m + 5 = 0\) using the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = 4\), \(c = 5\). Calculate the roots, which are \(m = -2 \pm i\).
03
Write the Complementary Solution
With roots \(-2 \pm i\), the complementary solution is \(y_c = e^{-2x}(C_1 \cos x + C_2 \sin x)\).
04
Form Particular Solution Trial
Since the non-homogeneous term is \(10e^{-3x}\), propose a trial particular solution: \(y_p = Ae^{-3x}\). Substitute \(y_p\) into the left side of the differential equation.
05
Differentiate the Trial Particular Solution
Calculate the derivatives: \(y_p = Ae^{-3x}, y_p' = -3Ae^{-3x}, y_p'' = 9Ae^{-3x}\).
06
Substitute and Solve for A
Substitute into the differential equation: \( (9Ae^{-3x} + 4(-3Ae^{-3x}) + 5Ae^{-3x}) = 10e^{-3x} \). Simplify to find \(A = -2\). The particular solution is \(y_p = -2e^{-3x}\).
07
Form General Solution
Combine the complementary and particular solutions: \(y = y_c + y_p = e^{-2x}(C_1 \cos x + C_2 \sin x) - 2e^{-3x}\).
08
Apply Initial Conditions
When \(x = 0, y = 4\), substitute to get: \(4 = C_1 - 2\). Also, \(y'(0) = 0\) implies another equation from the derivative of the general solution. Evaluate \(y'\) at \(x = 0\) to form the equation and solve.
09
Solve for Constants
Substituting the initial conditions \(y(0) = 4\) and \(y'(0) = 0\), solve the equations: \(4 = C_1 - 2\) leads to \(C_1 = 6\). From \(y'(0) = 0\), solve for \(C_2\) with the differentiated equation, resulting in \(C_2 = 3\).
10
Write the Particular Solution
Using the solved constants, the particular solution satisfying the initial conditions is: \(y = e^{-2x}(6 \cos x + 3 \sin x) - 2e^{-3x}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equation
A differential equation is a mathematical equation that relates some function with its derivatives. In our context, a differential equation involves an unknown function, often denoted as \(y(x)\), and its derivatives.
For our particular exercise, we have the differential equation \((D^2 + 4D + 5)y = 10e^{-3x}\). Here, \(D\) symbolizes the derivative \(\frac{d}{dx}\), and the goal is to find an expression for \(y\) that satisfies this equation.
For our particular exercise, we have the differential equation \((D^2 + 4D + 5)y = 10e^{-3x}\). Here, \(D\) symbolizes the derivative \(\frac{d}{dx}\), and the goal is to find an expression for \(y\) that satisfies this equation.
- The left-hand side, \(D^2 + 4D + 5\), is known as the differential operator, which acts on \(y\) to form derivatives of \(y\).
- The right-hand side, \(10e^{-3x}\), is known as the non-homogeneous term, and it influences the form of the particular solution we will find.
Initial Conditions
Initial conditions are additional pieces of information that allow us to find a particular solution to a differential equation. By specifying the value of the function and its derivative at a particular point, we can determine any constants in our solution.
For this exercise, the initial conditions given are \(x=0\), \(y=4\), and \(y'(0)=0\).
For this exercise, the initial conditions given are \(x=0\), \(y=4\), and \(y'(0)=0\).
- At \(x = 0\), the function \(y\) takes the value 4, meaning \(y(0) = 4\).
- The derivative \(y'\) is zero when \(x = 0\), so \(y'(0) = 0\).
Characteristic Equation
The characteristic equation emerges from the homogeneous part of a differential equation. It helps us find the complementary solution. To obtain the characteristic equation, replace derivatives with powers: for the operator \(D^2 + 4D + 5\), set \(D = m\) and solve the equation \(m^2 + 4m + 5 = 0\).
- This step transforms the differential operator into a polynomial in \(m\).
- Finding the roots of this polynomial yields the solutions relating to the homogeneous problem.
Quadratic Formula
The quadratic formula is the method we use to find the roots of a quadratic equation. When confronted with a quadratic in the form of \(ax^2 + bx + c = 0\), the roots can be calculated by:\[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Applying this to the characteristic equation \(m^2 + 4m + 5 = 0\):
- Here, \(a = 1\), \(b = 4\), and \(c = 5\).
- Plug these values into the quadratic formula. The discriminant, \(b^2 - 4ac\), is negative, leading to complex roots.
