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The integrating factor of $\frac{x d y}{d x}-y=x^{4}-3 x$ is
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Verified Answer
The correct answer is:
$\frac{1}{x}$
Since $x \frac{d y}{d x}-y=x^{4}-3 x$
$$
\therefore \quad \frac{d y}{d x}-\frac{y}{x}=x^{3}-3
$$
Hence
$$
I F=e^{\int P d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}
$$
$$
\therefore \quad \frac{d y}{d x}-\frac{y}{x}=x^{3}-3
$$
Hence
$$
I F=e^{\int P d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}
$$
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