Question
The general solution of the equation $$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0$$ is :
A.
$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
B.
$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$$
C.
$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$$
D.
$$x = 2 + k{e^{ - {{\tan }^{ - 1}}y}}$$
Answer :
$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
Solution :
$$\eqalign{
& \left( {1 + {y^2}} \right)\frac{{dx}}{{dy}} + x = {e^{{{\tan }^{ - 1}}y}} \cr
& {\text{or }}\,\frac{{dx}}{{dy}} + \frac{1}{{1 + {y^2}}}.x = \frac{{{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}\,\,{\text{which is in the linear form}}{\text{.}} \cr
& {\text{IF}} = {e^{\int {\frac{1}{{1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}.\,\,\,{\text{So, }}\,x.{e^{{{\tan }^{ - 1}}y}} = \int {\frac{{\left( {{e^{{{\tan }^{ - 1}}{y^2}}}} \right)dy}}{{1 + {y^2}}}} \cr
& \therefore x{e^{{{\tan }^{ - 1}}y}} = \frac{{{{\left( {{e^{{{\tan }^{ - 1}}y}}} \right)}^2}}}{2} + \frac{k}{2},\,\,\,\,\left( {{\text{putting }}{e^{{{\tan }^{ - 1}}y}} = z} \right) \cr} $$