The differential equation obtained by eliminating arbitrary constant from the equation $y^2=(x+c)^3$ is

The differential equation obtained by eliminating arbitrary constant from the equation $y^2=(x+c)^3$ is
  1. $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=27 y$
  2. $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3=-27 y$
  3. $8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3=27 y$
  4. $8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3+27 y=0$

Solution

$y^2=(x+c)^3$
Differentiating w.r.to $x$, we get $\begin{aligned} & 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=3(x+\mathrm{c})^2 \\ & \Rightarrow(x+\mathrm{c})^2=\frac{2 y}{3} \frac{\mathrm{~d} y}{\mathrm{~d} x} \\ & \Rightarrow(x+\mathrm{c})^6=\left(\frac{2 y}{3} \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3 \\ & \Rightarrow\left((x+\mathrm{c})^3\right)^2=\frac{8 y^3}{27}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3 \\ & \Rightarrow\left(y^2\right)^2=\frac{8 y^3}{27}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3 \\ & \Rightarrow y^4=\frac{8 y^3}{27}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3 \\ & \Rightarrow 27 y=8\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^3 \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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