Question

Solve the following differential equation $({\tan ^{ - 1}}y - x)dy = (1 + y)dx$

Answer 1

Differential equation is an equation that relates one or more functions and their derivatives. And an integrating factor is a function that is chosen to facilitate the solving of a given equation. The general for of differential equation is
$\dfrac{{dx}}{{dy}} + P(y)x = Q(y)$
$I.F = \,{e^{\int {P(y)dy} }}$

Stepwise solution
Given:
$({\tan ^{ - 1}}y - x)dy = (1 + {y^2})dx$
Stepwise solution:
$(1 + {y^2})dx\, = ({\tan ^{ - 1}}y - x)dy$
$\Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}} = \dfrac{x}{{1 + {y^2}}}$
$\Rightarrow \dfrac{{dx}}{{dy}} + \dfrac{x}{{1 + {y^2}}} = \dfrac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}}$
Hence,
$I.F = \,{e^{\int {P(y)dy} }}$
$= {e^{\int {\dfrac{1}{{1 + {y^2}}}dy} }}$
$I.\,F = {e^{{{\tan }^{ - 1}}y}}$
Hence, the above differential equation changes to
${e^{{{\tan }^{ - 1}}y}}\dfrac{{dy}}{{dx}} + \,\dfrac{{x{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}} = \dfrac{{{e^{{{\tan }^{ - 1}}y}}\,{{\tan }^{ - 1}}y}}{{1 + {y^2}}}$
$\Rightarrow \,{e^{{{\tan }^{ - 1}}y}}\,dx\, + \,\dfrac{{x{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}dy = \dfrac{{{e^{{{\tan }^{ - 1}}y}}\,{{\tan }^{ - 1}}y}}{{1 + {y^2}}}dy$
$\Rightarrow \,d({e^{{{\tan }^{ - 1}}y}}.\,x) = d({e^{{{\tan }^{ - 1}}y}})$
Integration of both the sides will result as
$\Rightarrow \,\int {d({e^{{{\tan }^{ - 1}}y}}\,x)} = \int {d({e^{{{\tan }^{ - 1}}y}})}$
$\Rightarrow {e^{{{\tan }^{ - 1}}y\,}}x = e{\,^{{{\tan }^{ - 1}}y}} + c$
$\Rightarrow \,x{e^{{{\tan }^{ - 1}}y}} - \,{e^{{{\tan }^{ - 1}}y}} + c$

Note:
The student must not forget to integrate and always remember to follow the general solution of differential equations.