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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20242 Marks
MCQ
$y=c^2+\frac{c}{x}$ is solution of  __________ .
  • ✓
    $x^4\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}-y=0$
  • B
    $x^2\left(\frac{d y}{d x}\right)^2+y=0$
  • C
    $x^3\left(\frac{d^2 y}{d x^2}\right)-x \frac{d y}{d x}+y=0$
  • D
    $x \frac{d^2 y}{d x^2}=4 y$
✓

Answer

Correct option: A.
$x^4\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}-y=0$
(A) $x^4\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}-y=0$
Explanation :
$y=c^2+\frac{c}{x}\ldots\ldots (1)$
$\begin{array}{ll}\therefore & \frac{d y}{d x}=-\frac{c}{x^2} \\
\therefore & c=-x^2 \cdot \frac{d y}{d x}\end{array}$
Substituting the value of $c$ in equation (1), we get
$\begin{array}{ll} y=\left(-x^2 \cdot \frac{d y}{d x}\right)^2+\frac{\left(-x^2 \cdot \frac{d y}{d x}\right)}{x} \\
\therefore \quad y=x^4\left(\frac{d y}{d x}\right)^2-x \cdot \frac{d y}{d x} \\
\therefore \quad x^4\left(\frac{d y}{d x}\right)^2-x \cdot \frac{d y}{d x}-y=0\end{array}$

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