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If the solution curve of the differential equation $\mathrm{\left(\left(\tan ^{-1} y\right)-x\right) d y=\left(1+y^{2}\right) d x}$ passes through the point $\mathrm{\left ( 1,0 \right )}$ , then the abscissa of the point on the curve whose ordinate is $\mathrm{\tan(1)}$ , is
Option: 1
$\mathrm{2e}$

Option: 2
$\mathrm{\frac{2}{e}}$

Option: 3
$\mathrm{2}$

Option: 4
$\mathrm{\frac{1}{e}}$

Answers (1)

best_answer

$\mathrm{\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{1} y}{1+y^{2}}}$

Linear differential equation

Integral factor

$\mathrm{\text { If }=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}} \\$

$\mathrm{\Rightarrow \quad x \times e^{\tan ^{-1} y}=\int \frac{\tan ^{-1} y e^{\tan ^{-1} y}}{1+y^{2}} d y}$

Let $\mathrm{\tan ^{-1} y=t \rightarrow x \times e^{\tan ^{-1} y}=\int \frac{t}{I} \frac{e^{t}}{II} d t}$

$\mathrm{\Rightarrow x \times e^{\tan ^{-1} y}=t e^{t}-\int e^{t} d t} \\$

$\mathrm{\Rightarrow x \times e^{\tan ^{-1} y}=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+c} \\$

$\mathrm{x=1, y=0 \Rightarrow 1\times e^{0}=0-e^{0}+c \Rightarrow c=2} \\$

$\mathrm{y=\tan 1 \Rightarrow x \times e^{1}=1 \times e^{1}-e^{1}+2 }\\$ $\Rightarrow \mathrm{x=2 / e}$

Hence the correct answer is option 2

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manish painkra

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