(1+x^2)dy/dx+y=e^tan^-1x answer

1.	(1+x^2)dy/dx+y=e^tan^-1x answer
Answer» $\large\underline{\sf{<klux>SOLUTION</klux>-}}$ The given Differential Equation is $\rm :\longmapsto\:(1+{x}^{2})\dfrac{dy}{dx} + y = {e}^{ {tan}^{ - 1} x}$ can be rewritten as $\rm :\longmapsto\:\dfrac{dy}{dx} +\dfrac{y}{1 + {x}^{2} } = \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }$ which is a Linear Differential equation. So, On comparing with, $\rm :\longmapsto\:\dfrac{dy}{dx} + py = q$ we get, $\red{\rm :\longmapsto\:p = \dfrac{1}{1 + {x}^{2} } } \\ \red{\rm :\longmapsto\:q = \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }}$ INTEGRATING Factor, I.F. is given by $\rm :\longmapsto\:I.F. = {e} \: ^{\displaystyle \int \: pdx}$ $\rm :\longmapsto\:I.F. = {e} \: ^{\displaystyle \int \: \dfrac{1}{1 + {x}^{2} } dx}$ $\bf\implies \:I.F. = {e}^{ {tan}^{ - 1} x}$ $\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \displaystyle \int \: \dfrac{dx}{1 + {x}^{2} } = {tan}^{ - 1} x + c \bigg \}}$ Solution of Linear Differential equation is $\rm :\longmapsto\:y \times I.F. = \displaystyle \int \:( q \times I.F. )dx$ $\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \displaystyle \int \sf \: {e}^{ {tan}^{ - 1} x} \times \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }dx$ $\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \displaystyle \int \sf \: \dfrac{{e}^{ {2tan}^{ - 1} x}}{1 + {x}^{2} }dx$ On RIGHT hand side of integral, $\red{\rm :\longmapsto\:Put \: {tan}^{ - 1}x = t} \\ \red{\rm :\longmapsto\:\dfrac{dt}{dx} = \dfrac{1}{1 + {x}^{2} }} \: \: \: \: \\ \red{\rm :\longmapsto\:\dfrac{dx}{1 + {x}^{2} } = dt} \: \: \: \: \: \:$ On substituting all these VALUES, we get $\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \displaystyle \int \sf \: {e}^{2t} dt$ $\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \dfrac{ {e}^{2t} }{2} + c$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \displaystyle \int \: {e}^{x}dx = {e}^{x} + c\bigg \}}$ $\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \dfrac{ {e}^{ {2tan}^{ - 1} x} }{2} + c$ $\bf :\longmapsto\:y \: \: = \dfrac{ {e}^{ {tan}^{ - 1} x} }{2} + c{e}^{ - {tan}^{ - 1} x}$

Answer»

$\large\underline{\sf{<klux>SOLUTION</klux>-}}$

The given Differential Equation is

$\rm :\longmapsto\:(1+{x}^{2})\dfrac{dy}{dx} + y = {e}^{ {tan}^{ - 1} x}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} +\dfrac{y}{1 + {x}^{2} } = \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }$

which is a Linear Differential equation.

So,

On comparing with,

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q$

we get,

$\red{\rm :\longmapsto\:p = \dfrac{1}{1 + {x}^{2} } } \\ \red{\rm :\longmapsto\:q = \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }}$

INTEGRATING Factor, I.F. is given by

$\rm :\longmapsto\:I.F. = {e} \: ^{\displaystyle \int \: pdx}$

$\rm :\longmapsto\:I.F. = {e} \: ^{\displaystyle \int \: \dfrac{1}{1 + {x}^{2} } dx}$

$\bf\implies \:I.F. = {e}^{ {tan}^{ - 1} x}$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \displaystyle \int \: \dfrac{dx}{1 + {x}^{2} } = {tan}^{ - 1} x + c \bigg \}}$

Solution of Linear Differential equation is

$\rm :\longmapsto\:y \times I.F. = \displaystyle \int \:( q \times I.F. )dx$

$\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \displaystyle \int \sf \: {e}^{ {tan}^{ - 1} x} \times \dfrac{{e}^{ {tan}^{ - 1} x}}{1 + {x}^{2} }dx$

$\rm :\longmapsto\:y \: {e}^{ {tan}^{ - 1} x} \: = \displaystyle \int \sf \: \dfrac{{e}^{ {2tan}^{ - 1} x}}{1 + {x}^{2} }dx$

On RIGHT hand side of integral,

$\red{\rm :\longmapsto\:Put \: {tan}^{ - 1}x = t} \\ \red{\rm :\longmapsto\:\dfrac{dt}{dx} = \dfrac{1}{1 + {x}^{2} }} \: \: \: \: \\ \red{\rm :\longmapsto\:\dfrac{dx}{1 + {x}^{2} } = dt} \: \: \: \: \: \:$