CUET · MATHS · PYQ PAPER 2023
For \(y=x e^x\), which of the following is correct ?
- A \(x \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-y=0\)
- B \(x \frac{d^2 y}{d x^2}-y \frac{d y}{d x}-x=0\)
- C \(x \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0\)
- D \(y \frac{d^2 y}{d x^2}-y \frac{d y}{d x}-x=0\)
Answer & Solution
Correct Answer
(A) \(x \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-y=0\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = 1 \cdot e^x + x \cdot e^x = e^x + xe^x\) \(\frac{d^2y}{dx^2} = e^x + (1 \cdot e^x + x \cdot e^x) = 2e^x + xe^x\) \(\frac{d^2y}{dx^2} - \frac{dy}{dx} = (2e^x + xe^x) - (e^x + xe^x) = e^x\) \(x \left( \frac{d^2y}{dx^2} - \frac{dy}{dx} \right) = xe^x\)…
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