Differential Equations — Class 12 Maths Solution

: The given equation is $(1 + {x^2})\cfrac{{dy}}{{dx}} + 2xy = \cfrac{1}{{1 + {x^2}}}$

$\Rightarrow \cfrac{{dy}}{{dx}} + \cfrac{{2x}}{{1 + {x^2}}}y = \cfrac{1}{{{{(1 + {x^2})}^2}}}$

…(1)
which is a linear equation of the type
Where, $P = \cfrac{{2x}}{{1 + {x^2}}}{\rm{ and }}Q = {\left( {\cfrac{1}{{1 + {x^2}}}} \right)^2}$

$\therefore$ $\int {Pdx} = \int {\cfrac{{2x}}{{1 + {x^2}}}dx} = \log |1 + {x^2}| = \log (1 + {x^2})$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\log \left( {1 + {x^2}} \right)}} = \left( {1 + {x^2}} \right)$

The solution is $y({\rm{I}}.{\rm{F}}.) = \int Q ({\rm{I}}.{\rm{F}}.)dx + C$

$\Rightarrow y \cdot (1 + {x^2}) = \int {\cfrac{{(1 + {x^2})}}{{{{(1 + {x^2})}^2}}}} dx + C$
$\Rightarrow y(1 + {x^2}) = {\tan ^{ - 1}}x + C$

…(2)
When $x = 1,y = 0,\therefore \,\,0 = {\tan ^{ - 1}}1 + C \Rightarrow C = - \cfrac{\pi }{4}$

Putting $C = - \cfrac{\pi }{4}$ in (2),

we get $y(1 + {x^2}) = {\tan ^{ - 1}}x - \cfrac{\pi }{4}$

which is the required solution.