$y=c^{2}+\frac{c}{x}$ is the solution of the differential equation

$y=c^{2}+\frac{c}{x}$ is the solution of the differential equation
  1. $x^{4}\left(\frac{d y}{d x}\right)^{2}+x\left(\frac{d y}{d x}\right)-y=0$
  2. $x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)-y=0$
  3. $x^{4}\left(\frac{d y}{d x}\right)^{2}-x\left(\frac{d y}{d x}\right)+y=0$
  4. $x^{4}\left(\frac{d y}{d x}\right)^{2}+x\left(\frac{d y}{d x}\right)+y=0$

Solution

Given $y=c^{2}+\frac{c}{x}$ Differentiating w.r.t. $x$ $\frac{d y}{d x}=0-\frac{c}{x^{2}}=-\frac{c}{x^{2}} \Rightarrow c=\left(-x^{2}\right)\left(\frac{d y}{d x}\right)$ Substituting value of $\mathrm{c}$ in given equation, We get $\begin{aligned} y &=\left[\left(-x^{2}\right)\left(\frac{d y}{d x}\right)\right]^{2}+\frac{\left(-x^{2}\right)\left(\frac{d y}{d x}\right)}{x} \\ &=x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x} \\ \therefore & x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}-y=0 \end{aligned}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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