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The solution of $\left(y-3 x^2\right) d x+x d y=0$ is
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Verified Answer
The correct answer is:
$y(x)=x^2+\frac{C}{x}$
We have,
$$
\begin{array}{rlrl}
& & \left(y-3 x^2\right) d x+x d y & =0 \\
\Rightarrow & y d x-3 x^2 d x+x d y & =0 \\
\Rightarrow & & y d x+x d y & =3 x^2 d x \\
\Rightarrow & & d x y & =3 x^2 d x
\end{array}
$$
On integrating both sides, we get
$$
x y=x^3+C \Rightarrow y=x^2+\frac{C}{x}
$$
$$
\begin{array}{rlrl}
& & \left(y-3 x^2\right) d x+x d y & =0 \\
\Rightarrow & y d x-3 x^2 d x+x d y & =0 \\
\Rightarrow & & y d x+x d y & =3 x^2 d x \\
\Rightarrow & & d x y & =3 x^2 d x
\end{array}
$$
On integrating both sides, we get
$$
x y=x^3+C \Rightarrow y=x^2+\frac{C}{x}
$$
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