Choose the correct answer from the given four option. The integra…

Question

Choose the correct answer from the given four option.
The integrating factor of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ is:

$\frac{\text{x}}{\text{e}^{\text{x}}}$
$\frac{\text{e}^{\text{x}}}{\text{x}}$
${\text{x}}\text{e}^{\text{x}}$
$\text{e}^{\text{x}}$

✓

Answer

$\frac{\text{e}^{\text{x}}}{\text{x}}$

Solution:

We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}}{\text{x}}-\text{y}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}-\text{xy}}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1}{\text{x}}+\frac{\text{y}(1-\text{x})}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\Big(\frac{1-\text{x}}{\text{x}}\Big)$

$\Rightarrow\text{y}=\frac{1}{\text{x}}$

Here, $\text{P}=\frac{-(1-\text{x})}{\text{x}},\text{Q}=\frac{1}{\text{x}}$

$\text{I.F.}=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{-\int\frac{1+\text{x}}{\text{x}}\text{dx}}=\text{e}^{\frac{\text{x}-1}{\text{x}}}$

$=\text{e}^{{\int\Big(1-\frac{1}{\text{x}}\Big)}\text{dx}}=\text{e}^{\int\text{x}-\log\text{x}}$

$=\text{e}^{\text{x}}.\text{e}^{\log\Big(\frac{1}{\text{x}}\Big)}=\text{e}^{\text{x}}.\frac{1}{\text{x}}$

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