Q.If $y = 3 \cos (\log x) + 4 \sin (\log x)$, show that $x^2 y_2 + xy_1 + y = 0$.

Concept: Homogeneous Linear Differential Equation (Cauchy-Euler type) — the given function is a combination of $\cos(\log x)$ and $\sin(\log x)$, which suggests a second-order ODE with variable coefficients.

Step 1: Differentiate $y = 3 \cos(\log x) + 4 \sin(\log x)$ with respect to $x$:

So $x y_1 = -3 \sin(\log x) + 4 \cos(\log x)$.

Step 2: Differentiate again:

Using the product rule:

Simplify:

$y_2 = -\frac{1}{x^2} \bigl( -3 \sin(\log x) + 4 \cos(\log x) \bigr) - \frac{1}{x^2} \bigl( 3 \cos(\log x) + 4 \sin(\log x) \bigr)$ …

We are given $y = 3 \cos (\log x) + 4 \sin (\log x)$. The task is to show that this function satisfies the differential equation $x^2 y_2 + xy_1 + y = 0$, where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$.

This is a classic homogeneous linear differential equation of the Cauchy-Euler type. The key insight: when the argument of trigonometric functions is $\log x$, derivatives produce factors of $\frac{1}{x}$, which then combine neatly with the $x^2$ and $x$ coefficients in the given expression. Let’s work through it step by step.

1. First derivative Differentiate $y$ with respect to $x$. Remember that $\frac{d}{dx}(\log x) = \frac{1}{x}$. Using the chain rule:

So:

2. Second derivative Differentiate $y_1$. Since $y_1$ is a product of $\frac{1}{x}$ and a function of $\log x$, use the product rule:

The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$.

For the second term, differentiate the bracket:

So the derivative of the bracket is:

Putting it all together:

Simplify:

Combine like terms:

3. Form the expression $x^2 y_2 + x y_1 + y$ Now substitute: $x^2 y_2 = x^2 \cdot \frac{1}{x^2} \left( -\sin(\log x) - 7 \cos(\log x) \right) = -\sin(\log x) - 7 \cos(\log x)$ …

Method: Verifying a Cauchy–Euler-Type Relation (Clear the Denominator First)

For a function like $y=A\cos(\log x)+B\sin(\log x)$ satisfying an equation of the form $x^2y_2+xy_1+y=0$, don't compute $y_2$ from scratch and then multiply by $x^2$ separately — instead, build $xy_1$ first (it's clean, with the $\frac{1}{x}$ cancelled), then differentiate that.

Steps

Step 1: Differentiate $y$ once to get $y_1$

Every $\log x$ term picks up a factor of $\frac{1}{x}$ from the chain rule.

Step 2: Multiply through by $x$ to get $xy_1$

This clears the $\frac{1}{x}$ and leaves a clean trig expression in $\log x$.

Step 3: Differentiate $xy_1$ using the product rule to get $y_1+xy_2$, then isolate $xy_2$ — multiply by $x$ once more for $x^2y_2$ …

Common Mistakes

Mistake 1: Forgetting the $\frac{1}{x}$ chain-rule factor on every $\log x$ term.

Why it's wrong: both $\cos(\log x)$ and $\sin(\log x)$ differentiate with an extra $\frac{1}{x}$ from the inner $\log x$ — dropping it on even one term throws off the whole cancellation. Correct approach: write $\frac{d}{dx}\log x=\frac{1}{x}$ explicitly at each differentiation step.

Mistake 2: Grouping the final $\sin(\log x)$/$\cos(\log x)$ coefficients incorrectly. …