2.7 Q-7

Question Statement

Given that

$$y = e^{a x} \sin(b x),$$

show that

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

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Background and Explanation

In this problem, we are asked to differentiate a product of an exponential and a trigonometric function. We need to use the product rule to compute both the first and second derivatives. Additionally, we will substitute the expressions for the first and second derivatives into the equation given in the problem to prove the result.

The basic differentiation rules we’ll use are:

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

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Solution

Step 1: Compute the first derivative

We start with:

$$y = e^{a x} \sin(b x).$$

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

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

Thus, the first derivative is:

$$\frac{dy}{dx} = e^{a x} (a \sin(b x) + b \cos(b x)).$$

Step 2: Compute the second derivative

Now, differentiate $\frac{dy}{dx} = e^{a x} (a \sin(b x) + b \cos(b x))$ again with respect to $x$. Use the product rule on each term:

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

Applying the product rule:

$$\frac{d^2 y}{dx^2} = e^{a x} \cdot \left[ a b \cos(b x) - b^2 \sin(b x) \right] + e^{a x} \cdot \left[ a^2 \sin(b x) + a b \cos(b x) \right].$$

Simplifying:

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

Step 3: Substitution into the equation

We now substitute the expressions for $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ into the given equation:

$$\frac{d^2 y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y.$$

From earlier, we know:

• $\frac{dy}{dx} = e^{a x} (a \sin(b x) + b \cos(b x))$,
• $\frac{d^2 y}{dx^2} = e^{a x} \left[ (a^2 + b^2) \sin(b x) + 2 a b \cos(b x) \right]$,
• $y = e^{a x} \sin(b x)$.

Now substitute these into the equation:

$$e^{a x} \left[ (a^2 + b^2) \sin(b x) + 2 a b \cos(b x) \right] - 2a \cdot e^{a x} (a \sin(b x) + b \cos(b x)) + (a^2 + b^2) e^{a x} \sin(b x).$$

Simplifying the terms:

$$e^{a x} \left[ (a^2 + b^2) \sin(b x) + 2 a b \cos(b x) - 2a (a \sin(b x) + b \cos(b x)) + (a^2 + b^2) \sin(b x) \right].$$

Notice the terms involving $\sin(b x)$ and $\cos(b x) will cancel out, and we are left with:

$$0.$$

Thus, we have shown that:

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

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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 find the first derivative $\frac{dy}{dx} = e^{a x} (a \sin(b x) + b \cos(b x))$.
2. Apply the product rule again to find the second derivative $\frac{d^2 y}{dx^2} = e^{a x} \left[ (a^2 + b^2) \sin(b x) + 2 a b \cos(b x) \right]$.
3. Substitute $\frac{dy}{dx}$, $\frac{d^2 y}{dx^2}$, and $y$ into the given equation.
4. Simplify the expression to show that the equation equals zero, proving the result.