2.7 Q-9

Question Statement

Given that

$$y = a \cos (\ln x) + b \sin (\ln x),$$

prove that

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

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Background and Explanation

This problem involves differentiating a function that combines both trigonometric and logarithmic functions. We will apply the chain rule for differentiation, as the argument of both the cosine and sine functions is $\ln x$, which requires special handling. We will differentiate $y$ twice to find $y_1$ (the first derivative) and $y_2$ (the second derivative), and then substitute these into the given equation to prove the result.

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Solution

Step 1: Find the first derivative $y_1$

Start with the given function:

$$y = a \cos (\ln x) + b \sin (\ln x).$$

Using the chain rule to differentiate, we get:

$$\frac{d y}{d x} = -a \sin (\ln x) \cdot \frac{1}{x} + b \cos (\ln x) \cdot \frac{1}{x}.$$

Thus, the first derivative is:

$$y_1 = \frac{d y}{d x} = \frac{-a \sin (\ln x) + b \cos (\ln x)}{x}.$$

Step 2: Multiply by $x$ to simplify the expression

Now, multiply both sides by $x$:

$$x \frac{d y}{d x} = -a \sin (\ln x) + b \cos (\ln x).$$

Step 3: Differentiate again to find the second derivative $y_2$

Next, we differentiate $x \frac{d y}{d x} = -a \sin (\ln x) + b \cos (\ln x)$ with respect to $x$ using the product rule:

$$\frac{d}{d x} \left( x \frac{d y}{d x} \right) = \frac{d}{d x} \left( -a \sin (\ln x) + b \cos (\ln x) \right).$$

Differentiating both sides:

$$\frac{d y}{d x} + x \frac{d^2 y}{d x^2} = -\frac{a \cos (\ln x)}{x} + \frac{-b \sin (\ln x)}{x}.$$

Step 4: Rearrange the expression

Rearrange the above equation to isolate $x^2 \frac{d^2 y}{dx^2}$ and $x \frac{d y}{dx}$:

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

Step 5: Substitute $y$ from the original equation

Note that from the original equation, we have:

$$y = a \cos (\ln x) + b \sin (\ln x).$$

Substitute this into the equation:

$$x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -y.$$

Step 6: Conclude the proof

Now, we can add $y$ to both sides:

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

Thus, we have proven the required identity.

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx} \left( \cos (\ln x) \right) = -\frac{\sin (\ln x)}{x}, \quad \frac{d}{dx} \left( \sin (\ln x) \right) = \frac{\cos (\ln x)}{x}.$

• Product Rule: $\frac{d}{dx} \left( x f(x) \right) = f(x) + x f'(x).$

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Summary of Steps

1. Differentiate $y = a \cos (\ln x) + b \sin (\ln x)$ to find the first derivative $y_1 = \frac{-a \sin (\ln x) + b \cos (\ln x)}{x}$.
2. Multiply the first derivative by $x$ to simplify it: $x \frac{d y}{d x} = -a \sin (\ln x) + b \cos (\ln x)$.
3. Differentiate again to find the second derivative $y_2$ using the product rule.
4. Rearrange the resulting equation to express it in terms of $y$.
5. Conclude that $x^2 \frac{d^2 y}{dx^2} + x \frac{d y}{dx} + y = 0$, as required.