Chapter 3: Problem 5
In Problems \(1-16\), the indicated function \(y_{1}(x)\) is a solution of the given equation. Use reduction of order or formula (5), as instructed, to find a second solution \(y_{2}(x)\). $$ y^{\prime \prime}-y=0 ; \quad y_{1}=\cosh x $$
Short Answer
Expert verified
The second solution is \( y_2(x) = \sinh x \).
Step by step solution
01
Rewrite the Differential Equation
The given differential equation is \( y'' - y = 0 \). We are told that \( y_1 = \cosh x \) is a solution.
02
Write the Second Solution
We assume the second solution \( y_2(x) \) is of the form \( y_2(x) = v(x) y_1(x) \). Therefore, \( y_2(x) = v(x) \cosh x \).
03
Derive the Derivatives
Find the first and second derivatives of \( y_2(x) = v(x) \cosh x \).The first derivative: \[ y_2' = v' \cosh x + v \sinh x \]The second derivative:\[ y_2'' = v'' \cosh x + 2v' \sinh x + v \cosh x \]
04
Substitute Derivatives into the Differential Equation
Substitute \( y_2', y_2'' \) into the differential equation \( y'' - y = 0 \):\[ v'' \cosh x + 2v' \sinh x + v \cosh x - (v \cosh x) = 0 \]This simplifies to:\[ v'' \cosh x + 2v' \sinh x = 0 \]
05
Simplify the Equation by Separating Variables
Divide by \( \cosh x \):\[ v'' + 2v' \tanh x = 0 \]This is a first-order separable differential equation for \( v' \).
06
Solve the First-order Differential Equation
Assume a solution of the form \( v' = w(x) \) and solve:\[ w' + 2w \tanh x = 0 \]The integrating factor is \( e^{\int 2 \tanh x \, dx} = (\cosh x)^2 \), so multiply through:\[ (w \cosh^2 x)' = 0 \]Thus, \( w \cosh^2 x = C \), leading to \( w = \frac{C}{\cosh^2 x} \).
07
Integrate to Find v(x)
Integrate \( w = \frac{C}{\cosh^2 x} \) to find \( v(x) \):\[ v(x) = \int \frac{C}{\cosh^2 x} \, dx = C \tanh x + D \]
08
Determine the Second Solution y_2(x)
Finally, substitute \( v(x) = C \tanh x + D \) into the assumed solution:\[ y_2(x) = (C \tanh x + D) \cosh x = C \sinh x + D \cosh x \] For a non-trivial second solution, we take \( C = 1, D = 0 \), giving:\[ y_2(x) = \sinh x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Differential Equations
Second-order differential equations are a type of differential equation that involve the second derivative of a function. These equations are prevalent in many fields, including physics and engineering, because they often describe dynamic systems like motion or waves.
The general form of a second-order linear differential equation is given by: \[ ay'' + by' + cy = f(x) \] where \( y'' \) is the second derivative, \( y' \) is the first derivative, \( y \) is the function, and \( a, b, \text{ and } c \) are constants.
In our specific problem, we consider the homogeneous differential equation: \[ y'' - y = 0 \] Here, the solution \( y_1 = \cosh x \) serves as one of the solutions to this equation.
The general form of a second-order linear differential equation is given by: \[ ay'' + by' + cy = f(x) \] where \( y'' \) is the second derivative, \( y' \) is the first derivative, \( y \) is the function, and \( a, b, \text{ and } c \) are constants.
In our specific problem, we consider the homogeneous differential equation: \[ y'' - y = 0 \] Here, the solution \( y_1 = \cosh x \) serves as one of the solutions to this equation.
- Homogeneous means \( f(x) \) is zero.
- The solution set is a linear combination of fundamental solutions.
- This form often appears in natural phenomena.
Hyperbolic Functions
Hyperbolic functions, like \( \cosh x \) and \( \sinh x \), are analogs of trigonometric functions but are based on hyperbolas instead of circles.
They are defined as follows:
In the context of the given exercise, the solution \( y_1 = \cosh x \) is given. The challenge lies in deducing another fundamental solution involving \( \sinh x \).
They possess identities similar to trigonometric functions, which aid in simplifying derivatives and integrals involving hyperbolic functions.
They are defined as follows:
- \( \cosh x = \frac{e^x + e^{-x}}{2} \)
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
In the context of the given exercise, the solution \( y_1 = \cosh x \) is given. The challenge lies in deducing another fundamental solution involving \( \sinh x \).
They possess identities similar to trigonometric functions, which aid in simplifying derivatives and integrals involving hyperbolic functions.
- \( \cosh^2 x - \sinh^2 x = 1 \)
- Derivative: \( \frac{d}{dx}(\sinh x) = \cosh x \)
- Derivative: \( \frac{d}{dx}(\cosh x) = \sinh x \)
Integration Techniques
Integration techniques are crucial when solving differential equations, especially when finding a function like \( v(x) \) in our exercise.
Some techniques used in the problem include:
Integration strategies are key in constructing solutions from derivatives, particularly in reduction of order problems.
Some techniques used in the problem include:
- Integration by Substitution: Useful when dealing with hyperbolic functions.
- Integration Factor: An advanced method used for first-order differential equations, such as \( v' + 2v \tanh x = 0 \).
- \( v(x) = \int \frac{C}{\cosh^2 x} \, dx = C \tanh x + D \)
Integration strategies are key in constructing solutions from derivatives, particularly in reduction of order problems.