The differential equation whose solution is $y=c^2+\frac{c}{x}$, where $c$ is constant, is

The differential equation whose solution is $y=c^2+\frac{c}{x}$, where $c$ is constant, is
  1. $x^4\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2-x \frac{\mathrm{d} y}{\mathrm{~d} x}-y=0$
  2. $x^2\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2+\frac{\mathrm{d} y}{\mathrm{~d} x}-y=0$
  3. $x\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2-x^2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$
  4. $x^4\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2-\frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$

Solution

$\begin{aligned} & y=c^2+\frac{c}{x} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-c}{x^2} \\ & \Rightarrow c=-x^2 \frac{\mathrm{d} y}{\mathrm{~d} x}\end{aligned}$ $\begin{aligned} & \text { Hence, } y=\left(-x^2 \frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2+\left(-x^2 \frac{\mathrm{d} y}{\mathrm{~d} x}\right) \times \frac{1}{x} \\ & \Rightarrow y=x^4 \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2-x \cdot \frac{\mathrm{d} y}{\mathrm{~d} x} \\ & \Rightarrow x^4\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2-x \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}-y=0\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 1)

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