If the solution curve of the differential equation

$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

<p>$\left( {({{\tan }^{ - 1}}y) - x} \right)dy = (1 + {y^2})dx$</p> <p>$${{dx} \over {dy}} + {x \over {1 + {y^2}}} = {{{{\tan }^{ - 1}}y} \over {1 + {y^2}}}$$</p> <p>$I.F. = {e^{\int {{1 \over {1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}$</p> <p>$\therefore$ Solution</p> <p>$$x.\,{e^{{{\tan }^{ - 1}}y}} = \int {{{{e^{{{\tan }^{ - 1}}y}}{{\tan }^{ - 1}}y} \over {1 + {y^2}}}dy} $$</p> <p>Let ${e^{{{\tan }^{ - 1}}y}} = t$</p> <p>${{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}} = dt$</p> <p>$= x{e^{{{\tan }^{ - 1}}y}} = \int {\ln tdt = t\ln t - t + c}$</p> <p>$\therefore$ $$ = x{e^{{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}}{\tan ^{ - 1}}y - {e^{{{\tan }^{ - 1}}y}} + c$$ ..... (i)</p> <p>$\because$ It passes through (1, 0) $\Rightarrow$ c = 2</p> <p>Now put y = tan1, then</p> <p>$ex = e - e + 2$</p> <p>$\Rightarrow x = {2 \over e}$</p>