Problem 7 - Differential Equations and Linear Algebra

Question

Verify that \(y(x)=e^{x} \sin x\) is a solution to the differential equation $$ 2 y \cot x-\frac{d^{2} y}{d x^{2}}=0 $$

Step-by-Step Solution

Answer

We found the first derivative of \(y(x) = e^x \sin x\) to be \(\frac{dy}{dx} = e^x(\cos x + \sin x)\) and the second derivative to be \(\frac{d^2y}{dx^2} = 2e^x\cos x\). Upon substituting these derivatives and \(y(x)\) into the given differential equation \(2y\cot x - \frac{d^2y}{dx^2} = 0\), we obtained \(2e^x\cos x - 2e^x\cos x = 0\), which is satisfied. Hence, we have verified that \(y(x) = e^x \sin x\) is a solution to the differential equation.

Step 1: Find the first and second derivatives of y(x)

To find the first derivative of \(y(x) = e^x \sin x\), we can apply the product rule: \[\frac{dy}{dx} = \frac{d}{dx} (e^x \sin x) = e^x \frac{d}{dx}\sin x + \sin x \frac{d}{dx} e^x\] Now, we know that the derivative of \(\sin x\) with respect to \(x\) is \(\cos x\), and the derivative of \(e^x\) with respect to \(x\) is \(e^x\), so: \[\frac{dy}{dx} = e^x (\cos x) + \sin x (e^x) = e^x(\cos x + \sin x)\] Next, we need to find the second derivative of \(y(x)\). It can be obtained by differentiating the first derivative with respect to \(x\). Again, we will use the product rule: \[\frac{d^2y}{dx^2} = \frac{d}{dx} \left(e^x(\cos x + \sin x)\right) = e^x \frac{d}{dx}(\cos x + \sin x) + (\cos x + \sin x) \frac{d}{dx} e^x\] The derivative of \(\cos x\) with respect to \(x\) is \(-\sin x\) and of \(\sin x\) is \(\cos x\), so: \[\frac{d^2y}{dx^2} = e^x(-\sin x + \cos x) + (\cos x + \sin x)e^x = e^x(-\sin x + \cos x + \cos x + \sin x)\] On simplifying the second derivative: \[\frac{d^2y}{dx^2} = 2e^x\cos x\]

Step 2: Plug in y(x) and its derivatives into the differential equation

Now, we will substitute the values of \(y\), \(\frac{dy}{dx}\), and \(\frac{d^2y}{dx^2}\) into the given differential equation \(2y\cot x - \frac{d^2y}{dx^2} = 0\) and see if the equation is satisfied: \[2(e^x \sin x)\cot x - \frac{d^2y}{dx^2} = 0\] Substitute the value of \(\frac{d^2y}{dx^2}\) that we found in Step 1: \[2(e^x \sin x)\cot x - 2e^x\cos x = 0\]

Step 3: Simplify the equation and check if it's satisfied

Let's simplify the equation further: \[2e^x\sin x \cot x - 2e^x\cos x = 0\] Since \(\cot x = \frac{\cos x}{\sin x}\), we can substitute this in the equation: \[2e^x\sin x \cdot \frac{\cos x}{\sin x} - 2e^x\cos x = 0\] The \(\sin x\) terms cancel out: \[2e^x\cos x - 2e^x\cos x = 0\] This equation is satisfied, i.e., the left-hand side is equal to the right-hand side (0). #Conclusion# We have successfully verified that \(y(x) = e^x \sin x\) is a solution to the given differential equation \(2y\cot x - \frac{d^2y}{dx^2} = 0\).

Key Concepts

Product Rule

The Product Rule is an essential tool in differentiation, especially when dealing with functions that are products of two or more sub-functions. If you have a function defined as the product of two functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of this product is given by:\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]This rule ensures that each component of the product is properly differentiated. In the context of our exercise, we applied the product rule to the function \( y(x) = e^x \sin x \). Here, \( u(x) = e^x \) and \( v(x) = \sin x \). Thus, when differentiating, one must compute:

• The derivative of the first function \( e^x \), which is simply \( e^x \).


• The derivative of the second function \( \sin x \), which is \( \cos x \).

By applying the product rule, we obtained the first derivative of \( y(x) \) as \( e^x (\cos x) + \sin x (e^x) = e^x(\cos x + \sin x) \). This reflects how effectively the product rule allows differentiation of complex expressions.

Second Derivative

The second derivative tells us how the rate of change of a function is itself changing. Calculating the second derivative involves differentiating the first derivative. In our exercise, this meant reinforcing the product rule.Continuing from the first derivative \( e^x(\cos x + \sin x) \), we need to apply the product rule again to find the second derivative. Thus, the process involved:

• Differentiating \( e^x \) to get \( e^x \), again.


• Finding the derivative of \( (\cos x + \sin x) \) which results in \( -\sin x + \cos x \).

Applying the product rule gave us: \[ \frac{d^2y}{dx^2} = e^x(-\sin x + \cos x) + (\cos x + \sin x)e^x = 2e^x \cos x \]This derivative is crucial for understanding the behavior of the original function, particularly when analyzing whether a function satisfies a differential equation, as done in the verification step.

Trigonometric Functions

Trigonometric functions, such as \( \sin x \) and \( \cos x \), are the backbone of trigonometry. They describe relationships in triangles and have various applications in calculus and differential equations. In our exercise, knowing their derivatives is key.For differentiation, remember:

• The derivative of \( \sin x \) is \( \cos x \).


• The derivative of \( \cos x \) is \( -\sin x \).

These derivatives are vital for solving our differential problem. Furthermore, the function \( \cot x \) appears as its trigonometric identity \( \cot x = \frac{\cos x}{\sin x} \) Helped simplify terms in the differential equation. Recognizing these relationships allows Algebraic simplifications, crucial in verifying the solution of the differential equation as zero balance indicates correctness.

Verification of Solutions

Verification of solutions involves confirming that a proposed function satisfies a given differential equation When plugging the function and its derivatives back into an equation, simplification should result in a correct statement, typically zero in our scenario.In the exercise, the solution process entailed:

• Substituting \( y(x) = e^x \sin x \) and its second derivative \( 2e^x \cos x \) into the differential equation \( 2y\cot x - \frac{d^2y}{dx^2} = 0 \).


• Simplifying the expression using the identity \( \cot x = \frac{\cos x}{\sin x} \), which helped cancel terms out.

Upon simplification, the expression reduced to a statement of equality: \[ 2e^x\cos x - 2e^x\cos x = 0 \]This confirms that each term resolves appropriately, thus proving \( y(x) = e^x \sin x \) is indeed a valid solution for the differential equation. This method underlines the importance of algebraic manipulation and understanding function behavior in validating solutions.