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Prove the following.
If y = x³ log(1/x), then prove that x(d² y)/(dx²) - 2(dy)/(dx) + 3x² = 0.
If $y = x^3 \log\left(\frac{1}{x}\right)$, then prove that $x\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 3x^2 = 0$.
Given: $y = x^3 \log\left(\frac{1}{x}\right)$
To show: $x\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 3x^2 = 0$
Answer
Answer
$y = x^3 \log\left(\frac{1}{x}\right) = -x^3 \log x$
$\frac{dy}{dx} = -3x^2 \log x - x^2$
$\frac{d^2 y}{dx^2} = -6x \log x - 3x - 2x = -6x \log x - 5x$
$x \frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 3x^2 = (-6x^2 \log x - 5x^2) - 2(-3x^2 \log x - x^2) + 3x^2$
$= -6x^2 \log x - 5x^2 + 6x^2 \log x + 2x^2 + 3x^2 = 0$
Hence proved.
- Step 1: Given $y = x^3 \log(1/x) = -x^3 \log x$.
- Step 2: Differentiating with respect to $x$: $\frac{dy}{dx} = -3x^2 \log x - x^2$.
- Step 3: Differentiating again: $\frac{d^2 y}{dx^2} = -6x \log x - 3x - 2x = -6x \log x - 5x$.
- Step 4: Multiplying by $x$: $x\frac{d^2 y}{dx^2} = -6x^2 \log x - 5x^2$.
- Step 5: Evaluating $-2\frac{dy}{dx} = 6x^2 \log x + 2x^2$.
- Step 6: Substituting into LHS: $x\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 3x^2 = (-6x^2 \log x - 5x^2) + (6x^2 \log x + 2x^2) + 3x^2 = 0$.