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The solution of $\frac{d y}{d x}=\left(\frac{x}{y}\right)^{-1 / 3}$ is
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Verified Answer
The correct answer is:
$y^{2 / 3}-x^{2 / 3}=c$
We have,
$$
\begin{gathered}
\frac{d y}{d x}=\left(\frac{x}{y}\right)^{-1 / 3} \\
\Rightarrow \quad y^{-1 / 3} d y=x^{-1 / 3} d x
\end{gathered}
$$
On integrating both sides, we get
$$
\begin{aligned}
& \frac{y^{-1 / 3+1}}{-\frac{1}{3}+1}=\frac{x^{-1 / 3+1}}{-\frac{1}{3}+1}+c^{\prime} \\
& \Rightarrow \quad \frac{y^{2 / 3}}{2 / 3}=\frac{x^{2 / 3}}{2 / 3}+c^{\prime} \\
& \Rightarrow \quad y^{2 / 3}-x^{2 / 3}=\frac{2}{3} c^{\prime}=c \\
&
\end{aligned}
$$
$$
\begin{gathered}
\frac{d y}{d x}=\left(\frac{x}{y}\right)^{-1 / 3} \\
\Rightarrow \quad y^{-1 / 3} d y=x^{-1 / 3} d x
\end{gathered}
$$
On integrating both sides, we get
$$
\begin{aligned}
& \frac{y^{-1 / 3+1}}{-\frac{1}{3}+1}=\frac{x^{-1 / 3+1}}{-\frac{1}{3}+1}+c^{\prime} \\
& \Rightarrow \quad \frac{y^{2 / 3}}{2 / 3}=\frac{x^{2 / 3}}{2 / 3}+c^{\prime} \\
& \Rightarrow \quad y^{2 / 3}-x^{2 / 3}=\frac{2}{3} c^{\prime}=c \\
&
\end{aligned}
$$
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