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Question: Answered & Verified by Expert
The solution of $\left(y-3 x^2\right) d x+x d y=0$ is
MathematicsDifferential EquationsJEE Main
Options:
  • A $y(x)=\sin x+\frac{1}{x^2}+C$
  • B $y(x)=\cos x-\frac{1}{x^2}+C$
  • C $y(x)=x^2+\frac{C}{x}$
  • D $y(x)=\sqrt{x}+\frac{C}{x}$
Solution:
1697 Upvotes 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
$$
x y=x^3+C \Rightarrow y=x^2+\frac{C}{x}
$$

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