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The integrating factor of \( \frac{d y}{d x}+y=\frac{1+y}{x} \) is
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Verified Answer
The correct answer is:
\( \frac{e^{x}}{x} \)
Given differential equation,
\[
\begin{array}{l}
\frac{d y}{d x}+y=\frac{1+y}{x} \\
\Rightarrow \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x}
\end{array}
\]
So, integrating factor is given by
I.F. \( =e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x} e^{-\ln x}=\frac{e^{x}}{x} \)
\[
\begin{array}{l}
\frac{d y}{d x}+y=\frac{1+y}{x} \\
\Rightarrow \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x}
\end{array}
\]
So, integrating factor is given by
I.F. \( =e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x} e^{-\ln x}=\frac{e^{x}}{x} \)
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