The differential equation obtained by eliminating the arbitrary constants from the equation $y^{2}=(2…
The differential equation obtained by eliminating the arbitrary constants from the
equation $y^{2}=(2 x+c)^{5}$ is
$\left(\frac{d y}{d x}\right)^{4}-625 y^{4}=0$
$\left(\frac{d y}{d x}\right)^{5}-3125 y^{3}=0$
$\left(\frac{d y}{d x}\right)^{3}-125 y^{3}=0$
$x y \frac{d y}{d x}=5$
Solution
We have $y^{2}=(2 x+c)^{5}$ ...(i)
$\therefore 2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=5(2 \mathrm{x}+\mathrm{c})^{4}(2) \Rightarrow \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=5(2 \mathrm{x}+\mathrm{c})^{4}$
$\therefore(2 x+c)=\left[\frac{y}{5}\left(\frac{d y}{d x}\right)\right]^{\frac{1}{4}}$ and substituting value of $(2 x+c)$ in eq. (i), we write
$y^{2}=\left[\frac{y}{5}\left(\frac{d y}{d x}\right)\right]^{\frac{5}{4}}$
Raising both sides to power of 4 , we get
$\begin{aligned}
& y^{8}=\left[\frac{y}{5}\left(\frac{d y}{d x}\right)\right]^{5}=\frac{y^{5}}{3125}\left(\frac{d y}{d x}\right)^{5} \\
\therefore & 3125 y^{3}=\left(\frac{d y}{d x}\right)^{5}
\end{aligned}$