Let the solution curve $y=y(x)$ of the differential equation

$$ \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : } $$

<p>$${{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}$$</p> <p>$$IF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}$$</p> <p>Let ${\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt$</p> <p>$$ \Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}$$</p> <p>$$ \Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}$$</p> <p>$\therefore$ Solution is</p> <p>$$y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c} $$</p> <p>$$ \Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c$$</p> <p>$y(0) = 0 \Rightarrow c = 0$</p> <p>$x = 1$</p> <p>$y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1$</p> <p>$\Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}$</p> <p>$\Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}$</p>