2.7 Q-6

Question Statement

Given that

$$y = e^x \sin x,$$

show that

$$\frac{d^2 y}{dx^2} = 2 \frac{dy}{dx} + 2y = 0.$$

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Background and Explanation

This problem involves finding the second derivative of a function that is a product of an exponential and trigonometric function. We will need to apply the product rule for differentiation twice. Additionally, we will use the result of the first derivative to help compute the second derivative. Understanding these basic differentiation rules is crucial for solving the problem.

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Solution

Step 1: Compute the first derivative

We are given that

$$y = e^x \sin x.$$

To find the first derivative $\frac{dy}{dx}$, we apply the product rule:

$$\frac{dy}{dx} = \frac{d}{dx} \left( e^x \sin x \right) = e^x \sin x + e^x \cos x.$$

Thus, the first derivative is:

$$\frac{dy}{dx} = e^x \sin x + e^x \cos x.$$

Step 2: Compute the second derivative

Next, differentiate $\frac{dy}{dx} = e^x \sin x + e^x \cos x$ with respect to $x$ to find $\frac{d^2y}{dx^2}$. Again, apply the product rule to each term:

$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( e^x \sin x \right) + \frac{d}{dx} \left( e^x \cos x \right).$$

This gives:

$$\frac{d^2 y}{dx^2} = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x.$$

Step 3: Simplify the second derivative

Simplify the expression for $\frac{d^2 y}{dx^2}$:

$$\frac{d^2 y}{dx^2} = 2 e^x \cos x.$$

Step 4: Substitute into the desired equation

We know that

$$\frac{dy}{dx} = e^x \sin x + e^x \cos x,$$

so we can substitute this into the equation for $\frac{d^2 y}{dx^2}$. We get:

$$\frac{d^2 y}{dx^2} = 2 \left( \frac{dy}{dx} - e^x \sin x \right).$$

Thus,

$$\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0.$$

This is the required result, and we have proven that:

$$\frac{d^2 y}{dx^2} = 2 \frac{dy}{dx} + 2y = 0.$$

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Key Formulas or Methods Used

• Product Rule for differentiation:

$$\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x).$$

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Summary of Steps

1. Apply the product rule to differentiate $y = e^x \sin x$ to get $\frac{dy}{dx} = e^x \sin x + e^x \cos x$.
2. Differentiate $\frac{dy}{dx}$ again using the product rule to get $\frac{d^2 y}{dx^2} = 2 e^x \cos x$.
3. Substitute the expression for $\frac{dy}{dx}$ into $\frac{d^2 y}{dx^2} = 2 \left( \frac{dy}{dx} - e^x \sin x \right)$.
4. Rearrange to show $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$.