Question
The solution of the differential equation $$\frac{{dy}}{{dx}} + \frac{{2yx}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}{\text{ is :}}$$
A.
$$y\left( {1 + {x^2}} \right) = c + {\tan ^{ - 1}}x$$
B.
$$\frac{y}{{1 + {x^2}}} = c + {\tan ^{ - 1}}x$$
C.
$$y\,\log \left( {1 + {x^2}} \right) = c + {\tan ^{ - 1}}x$$
D.
$$y\left( {1 + {x^2}} \right) = c + {\sin ^{ - 1}}x$$
Answer :
$$y\left( {1 + {x^2}} \right) = c + {\tan ^{ - 1}}x$$
Solution :
Given differential equation is $$\frac{{dy}}{{dx}} + \frac{{2yx}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$$ which is a linear.
Differential equation of the form: $$\frac{{dy}}{{dx}} + Py = Q$$
On comparing, we have
$$\eqalign{
& P = \frac{{2x}}{{1 + {x^2}}}{\text{ and }}Q = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }} = {e^{\log \left( {1 + {x^2}} \right)}} = \left( {1 + {x^2}} \right) \cr} $$
$$\therefore $$ Solution is
$$\eqalign{
& y\left( {1 + {x^2}} \right) = \int {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}} \left( {1 + {x^2}} \right)dx + c \cr
& \Rightarrow y\left( {1 + {x^2}} \right) = \int {\frac{1}{{\left( {1 + {x^2}} \right)}}} dx + c \cr
& \Rightarrow y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x + c \cr} $$