The integrating factor of the differential equation $\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3$ is

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Aicharya · Accepted Answer

$\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \quad$ is linear differentiatial equation $	herefore 	ext { I.F. }=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}=\mathrm{e}^{\log x}=\mathrm{x}$

The integrating factor of the differential equation $\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3$ is

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