Problem 5 - Advanced Engineering Mathematics

Question

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 \(y_{2}(x)\). $$ y^{\prime \prime}-y=0 ; \quad y_{1}=\cosh x $$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = \sinh x \).

Step 1: Rewrite the Differential Equation

The given differential equation is \( y'' - y = 0 \). We are told that \( y_1 = \cosh x \) is a solution.

Step 2: 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, \( y_2(x) = v(x) \cosh x \).

Step 3: 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:\[ y_2'' = v'' \cosh x + 2v' \sinh x + v \cosh x \]

Step 4: 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:\[ v'' \cosh x + 2v' \sinh x = 0 \]

Step 5: Simplify the Equation by Separating Variables

Divide by \( \cosh x \):\[ v'' + 2v' \tanh x = 0 \]This is a first-order separable differential equation for \( v' \).

Step 6: 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 \( w = \frac{C}{\cosh^2 x} \).

Step 7: Integrate to Find v(x)

Integrate \( w = \frac{C}{\cosh^2 x} \) to find \( v(x) \):\[ v(x) = \int \frac{C}{\cosh^2 x} \, dx = C \tanh x + D \]

Step 8: 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:\[ y_2(x) = \sinh x \]

Key Concepts

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.

• Homogeneous means \( f(x) \) is zero.
• The solution set is a linear combination of fundamental solutions.
• This form often appears in natural phenomena.

The task is to find a second solution \( y_2(x) \) using the method of reduction of order.

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:

• \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( \sinh x = \frac{e^x - e^{-x}}{2} \)

These functions often appear in solutions to differential equations due to their similarity in behavior to such functions over real numbers.
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: \( \frac{d}{dx}(\cosh x) = \sinh x \)

Understanding these functions facilitates solving the differential equation for the second solution.

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 by Substitution: Useful when dealing with hyperbolic functions.
• Integration Factor: An advanced method used for first-order differential equations, such as \( v' + 2v \tanh x = 0 \).

The integration factor technique is highlighted through multiplying the differential equation by \((\cosh x)^2\). This simplifies it into a form that's easy to integrate:\[ (w \cosh^2 x)' = 0 \]This allows us to establish that \( w \cosh^2 x = C \), and later use integration techniques to find \( v(x) \):

• \( v(x) = \int \frac{C}{\cosh^2 x} \, dx = C \tanh x + D \)

Choosing specific constants like \( C = 1 \) and \( D = 0 \) then guides us to the second solution \( \sinh x \).
Integration strategies are key in constructing solutions from derivatives, particularly in reduction of order problems.