GATE 2024 BT – Question 64
Let $y(x) = x^2\ln x$ for $x > 0$, be a solution of $x^2\frac{d^2y}{dx^2} + 4y = \alpha x\frac{dy}{dx}$. Then the value of $\alpha$ is ________.
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Correct answer: 3
Explanation
$y' = 2x\ln x + x$ and $y'' = 2\ln x + 3$. Left side: $2x^2\ln x + 3x^2 + 4x^2\ln x = 6x^2\ln x + 3x^2$. Right side: $\alpha(2x^2\ln x + x^2)$. Matching coefficients gives $\alpha = 3$.