The Solution of differential equation dx + 1 + x 2 = ( 1 + x 2 ) 2 is

Given that dx + 1 + x 2 = ( 1 + x 2 ) 2 Here, P = 1 + x 2 and Q = ( 1 + x 2 ) 2 which is a linear differential equation. = e 1 + x 2 dx Put 1 + x 2 = t 2xdx = dt = e t = e t = e ( 1 + x 2 ) = 1 + x 2 The general Solution is y ( 1 + x 2 ) = ) ( 1 + x 2 ) 2 + C y ( 1 + x 2 ) = 1 + x 2 dx + C y ( 1 + x 2 ) = - 1 x + C

Differential Equations — Class 12 Maths Solution

exemplar objective MCQ NCERT EXEMP.Q.75,Page.201

Question

The Solution of differential equation

$\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$ is

(a) $y\left( {1 + {x^2}} \right) = C + {\tan ^{ - 1}}x$ ✓ Correct
(b) $\frac{y}{{1 + {x^2}}} = C + {\tan ^{ - 1}}x$
(c) $y\log \left( {1 + {x^2}} \right) = C + {\tan ^{ - 1}}x$
(d) $y\left( {1 + {x^2}} \right) = C + {\sin ^{ - 1}}x$

Step-by-step Solution

Given that

$\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$

Here, $P = \frac{{2x}}{{1 + {x^2}}}$
and $Q = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$

which is a linear differential equation.
$\therefore$

${\rm{IF}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}} dx}}$

Put $1 + {x^2} = t \Rightarrow 2xdx = dt$

$\therefore {\rm{IF}} = {e^{\int {\frac{{dt}}{t}} }} = {e^{\log t}} = {e^{\log \left( {1 + {x^2}} \right)}} = 1 + {x^2}$

The general Solution is
$y \cdot \left( {1 + {x^2}} \right) = \int {\left( {1 + {x^2}} \right)} \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} + C$

$\Rightarrow$ $y\left( {1 + {x^2}} \right) = \int {\frac{1}{{1 + {x^2}}}} dx + C$

$\Rightarrow$ $y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x + C$

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