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  • The integrating factor of the differential equation \frac{d y}{d x}+y=\frac{1+y}{x} is (a) \frac{x}{e^{x}}

The integrating factor of the differential equation \frac{d y}{d x}+y=\frac{1+y}{x} is
(a) \frac{x}{e^{x}}
(b) \frac{e^{x}}{x}
(c) x e^{x}
(d) e^{x}

Answers (1)

The answer is the option (b) \frac{e^{x}}{x}

Explanation: -

\Rightarrow \frac{d y}{d x}=\frac{1}{x}+\frac{y(1-x)}{x}
\Rightarrow \quad \frac{d y}{d x}-\left(\frac{1-x}{x}\right) y=\frac{1}{x}
This is a linear differential equation.
On comparing it with \frac{d y}{d x}+P y=Q, we get
\\ P=\frac{-(1-x)}{x}, Q=\frac{1}{x} \\ \mathrm{IF},=\int P d x=e^{-\int \frac{1-x}{x} d x} = \frac{e^x}{x}
 

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