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Magazine Chapter 4: Problem 8
In Problems 1-24 find a second solution of each differential equation. Use reduction of order or formula (5) as instructed. Assume an appropriate interval of validity. $$ y^{\prime \prime}-25 y=0 ; \quad y_{1}=e^{5 x} $$
Short Answer
Expert verified
The second solution is \( y_2 = e^{-5x} \).
Step by step solution
01
Understanding the Problem
We need to find a second linearly independent solution to the differential equation: \( y'' - 25y = 0 \). We are given one solution \( y_1 = e^{5x} \).
02
Apply the Reduction of Order Method
Since we have one solution \( y_1 \), we use the reduction of order technique. Assume a solution of the form \( y_2 = v(x) y_1 = v(x) e^{5x} \) where \( v(x) \) is a new function to be determined.
03
Derivatives of the Assumed Solution
Calculate the first derivative: \( y_2' = v'(x) e^{5x} + 5v(x) e^{5x} \). Calculate the second derivative: \( y_2'' = v''(x) e^{5x} + 10v'(x) e^{5x} + 25v(x) e^{5x} \).
04
Substitute back into Differential Equation
Substitute \( y_2'' \) and \( y_2 \) back into the original differential equation: \( (v''(x) e^{5x} + 10v'(x) e^{5x} + 25v(x) e^{5x}) - 25(v(x) e^{5x}) = 0 \). This simplifies to \( v''(x) e^{5x} + 10v'(x) e^{5x} = 0 \).
05
Simplify and Solve for \( v(x) \)
Divide the entire equation by \( e^{5x} \): \( v''(x) + 10v'(x) = 0 \). This is a first-order linear differential equation in terms of \( v'(x) \), let \( u = v'(x) \), then \( u' = v''(x) \). It becomes \( u' + 10u = 0 \). Solve this by separation of variables.
06
Solve the First-Order Linear Equation
The equation \( u' + 10u = 0 \) can be rewritten and solved by separation: \( \frac{du}{dx} = -10u \) leading to \( \frac{du}{u} = -10 dx \). Integrating both sides gives \( \ln|u| = -10x + C \). Thus \( u = v'(x) = Ce^{-10x} \).
07
Integrate to Find \( v(x) \)
Integrate \( v'(x) = Ce^{-10x} \): \( v(x) = \int Ce^{-10x} dx = -\frac{C}{10} e^{-10x} + K \).
08
Write the Solution \( y_2(x) \)
Combine \( v(x) \) with \( y_1 \) to find \( y_2 \): \( y_2(x) = v(x) e^{5x} = (-\frac{C}{10} e^{-10x} + K)e^{5x} = -\frac{C}{10} e^{-5x} + Ke^{5x} \). Since we seek a linearly independent solution, consider \( y_2 = e^{-5x} \) by setting constants appropriately.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reduction of Order
The reduction of order method is a powerful technique used to find a second solution to a second-order linear homogeneous differential equation when one solution is already known. In the context of the problem, we are tasked with determining another solution to the equation: \( y'' - 25y = 0 \), given that the first solution \( y_1 = e^{5x} \) is known. Reduction of order exploits the known solution to create a second solution that is linearly independent. Here's a step-by-step breakdown:
- Assume the second solution \( y_2 \) takes the form \( y_2 = v(x) y_1 = v(x) e^{5x} \), where \( v(x) \) is a function to be determined.
- The challenge is then to find \( v(x) \) such that when \( y_2 \) is placed into the differential equation, it still satisfies the equation.
- This involves finding the derivatives of \( y_2 \) and substituting them back into the differential equation.
Linearly Independent Solutions
Understanding linearly independent solutions is essential in the context of solving differential equations. Two solutions \( y_1 \) and \( y_2 \) of a homogeneous differential equation are said to be linearly independent if their combination cannot generate one another. Mathematically, this involves checking that the Wronskian is non-zero:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} eq 0\]For our equation \( y'' - 25y = 0 \), where \( y_1 = e^{5x} \), we wish to find another solution \( y_2 = e^{-5x} \) that is linearly independent.
- Ensure both solutions solve the differential equation and cannot be expressed as simple scalar multiples of each other.
- This guarantees that the general solution to the differential equation, structured as a linear combination of \( y_1 \) and \( y_2 \), covers all possibilities.
Homogeneous Differential Equations
Homogeneous differential equations are a particular type of differential equation where the system itself equals zero. Their general form can be written as:\[\text{Linear homogeneous differential equation: } a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = 0\]In our specific problem, the equation is:\( y'' - 25y = 0 \)This is a second-order linear homogeneous differential equation with constant coefficients. Key characteristics include:
- They do not have any "external" or "forcing" function (the right-hand side equals zero).
- The solutions form a vector space allowing us to use superposition. This means any linear combination of solutions is also a solution.
Second-Order Differential Equations
Second-order differential equations involve the second derivative of an unknown function, typically modeling systems with acceleration, curvature, or bending (for example, springs, beams, circuits). The standard form is:\( a(x)y'' + b(x)y' + c(x)y = f(x) \)Our example \( y'' - 25y = 0 \) is among the simplest, having constant coefficients and being homogeneous. Important concepts of such equations include:
- Characteristic Equation: Derived from turning the differential equation into a polynomial, helping find exponential solutions such as \( e^{rx} \).
- Particular and General Solutions: The general solution is a combination of all particular solutions, covering the entire solution space permissible.
- Applications: These often relate to wave motion, electrical circuits, and mechanical systems, where natural modes of vibration and response are of interest.
