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Integrating factor of the differential equation $(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^2$, is |
$-(x+1)$ $\log (x+1)$ $e^{x+1}$ $\frac{1}{x+1}$ |
$\frac{1}{x+1}$ |
We have, $\frac{d y}{d x}-\frac{1}{x+1} y=e^{3 x}(x+1)$ ∴ Integrating factor = $e^{-\int \frac{1}{x+1} d x}=e^{-\log (x+1)}=\frac{1}{x+1}$ |
