Problem 6 - Advanced Engineering Mathematics

Question

In Problems 5 and 6, compute \(y^{\prime}\) and \(y^{\prime \prime}\) and then combine these derivatives with \(y\) as a linear second-order differential equation that is free of the symbols \(c_{1}\) and \(c_{2}\) and has the form \(F\left(y, y^{\prime}, y^{\prime \prime}\right)=0\). The symbols \(c_{1}\) and \(c_{2}\) represent constants. $$ y=c_{1} e^{x} \cos x+c_{2} e^{x} \sin x $$

Step-by-Step Solution

Answer

The second-order differential equation is \( y'' - 2y' + 2y = 0 \).

Step 1: Find First Derivative

To find the first derivative of the given function \( y = c_{1} e^{x} \cos x + c_{2} e^{x} \sin x \), apply the product rule. Recall that the derivative of \( e^{x} \) with respect to \( x \) is \( e^{x} \), and use the chain rule for \( \cos x \) and \( \sin x \). The first derivative is:\[y^{\prime} = c_{1} (e^{x} \cos x \cdot 1 + (-e^{x}) \sin x) + c_{2} (e^{x} \sin x \cdot 1 + e^{x} \cos x) \]This simplifies to:\[y^{\prime} = e^{x}(c_{1}(\cos x - \sin x) + c_{2}(\sin x + \cos x))\]

Step 2: Find Second Derivative

Next, find the second derivative \( y^{\prime \,\prime} \) by differentiating \( y^{\prime} \ = e^{x}(c_{1}(\cos x - \sin x) + c_{2}(\sin x + \cos x)) \ \) using the same rules. This gives us:\[y^{\prime \,\prime} = e^{x}[c_{1}((-\sin x) - (-\cos x)) + (\cos x - \sin x)] + c_{2}[(\cos x) + (\sin x) + (\sin x + \cos x)]\]Simplifying:\[y^{\prime \,\prime} = e^{x}(c_{1}(-2\sin x) + c_{2}(2\cos x))\]

Step 3: Set Up Linear Second-order Differential Equation

Using \( y \), \( y^{\prime} \), and \( y^{\prime \,\prime} \), we form the differential equation by eliminating \( c_{1} \) and \( c_{2} \). We hypothesize a linear combination:\[F(y, y^{\prime}, y^{\prime \prime}) = Ay + By^{\prime} + Cy^{\prime \prime} = 0\]By substituting our expressions for \( y \), \( y^{\prime} \), and \( y^{\prime \prime} \) into this equation, and choosing coefficients such that it is free of \( c_{1} \) and \( c_{2} \), we find:\[y^{\prime \prime} - 2y^{\prime} + 2y = 0\]

Step 4: Validate and Simplify

Ensure all expressions are consistent and check through substitution:Substituting back each \( y \), \( y^{\prime} \), \( y^{\prime \prime} \) confirms that:\[A = 2, B = -2, C = 1\]Thus, the derived equation satisfies \( y^{\prime \prime} - 2y^{\prime} + 2y = 0 \) with no \( c_1 \) and \( c_2 \).

Key Concepts

Derivatives

To find the derivatives of a function means to calculate how the function changes as its input changes. In math terms, the derivative is a way to represent the slope of a function at any given point. For our problem, the function involves elements like the exponential function, denoted as \( e^x \), and trigonometric functions, like \( \cos x \) and \( \sin x \). Each of these has its specific rules for differentiating:

• The derivative of \( e^x \) with respect to \( x \) is simply \( e^x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \sin x \) is \( \cos x \).

When we differentiate the original function \( y = c_1 e^x \cos x + c_2 e^x \sin x \), we apply these rules to find the first and second derivatives, useful for our linear second-order differential equation.

Product Rule

The product rule is essential when you are dealing with derivatives of products of functions. Imagine you have two functions multiplied together, like \( e^x \) and \( \cos x \), which appear in our original equation. The product rule helps you find their derivative by following this formula:
\[(uv)' = u'v + uv'\]Where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their derivatives. Applying the product rule to our function \( y = c_1 e^x \cos x + c_2 e^x \sin x \) means:

• Compute the derivative of the first function in the product (e.g., \( e^x \) in the first term).
• Multiply it with the second unchanged function (e.g., \( \cos x \)).
• Add the product of the first unchanged function (e.g., \( e^x \)) and the derivative of the second function (e.g., \( -\sin x \)).

These steps apply to each term in the expression where functions are multiplied.

Chain Rule

The chain rule is a powerful tool for differentiating composite functions, meaning functions within functions. In this exercise, we apply the chain rule alongside the product rule to effectively find derivatives.
The formula for the chain rule is:
\[\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)\]This tells us to differentiate the outer function, evaluate it at the inner function, and then multiply it by the derivative of the inner function. In our context, when you encounter terms like \( e^x \cos x \) or \( e^x \sin x \), you handle each part separately using the chain rule. If a function appears inside another, always keep handy:

• Differentiating \( \cos x \) leads to \( -\sin x \).
• Differentiating \( \sin x \) leads to \( \cos x \).

The rule helps us link these derivatives correctly in the context of our composite functions.

Differential Equation Solutions

In the exercise, the goal is to form and solve a linear second-order differential equation. These are equations that involve up to the second derivative of a function. To solve them, we often aim for an equation of the form \( F(y, y', y'') = 0 \).
Once we find both the first and second derivatives, the next step is to eliminate constants, \( c_1 \) and \( c_2 \). We achieve this by cleverly setting up the equation such that these constants cancel out:

• Use linear combinations of \( y \), \( y' \), and \( y'' \) to eliminate constants.
• Simplify the resulting equation to derive an expression free of the original function's constants.

Eventually, the exercise demonstrates how these steps result in a tidy equation, \( y'' - 2y' + 2y = 0 \). This form satisfies the criteria of being independent of the constants and provides a beautiful differential equation solution.