Discussion Forum

 The solution of differential equation $(1+y^{2})+(x-e^{tan^{-1}y})\frac{dy}{dx}=0$

Answer:

Explanation:

$1+y^{2}+(x-e^{\tan^{-1}y})\frac{\text{d}y}{\text{d}x}=0$

$\Rightarrow (1+y^{2}) dx=(e^{\tan^{-1}y}-x)dy$

 $\Rightarrow \frac{dy}{dx}=\frac{e^{\tan^{-1}y}-x}{1+y^{2}}$

$\Rightarrow \frac{dy}{dx}+\frac{1}{1+y^{2}}.x=\frac{e^{\tan^{-1}y}}{1+y^{2}}$

 Which is the linear different equation of the `

form  $\frac{dx}{dy}+Rx=S$ , where R and S are functions of y or constant (s)

$\therefore$ I.F=$e^{\int_{}^{}\frac{1}{1+y^{2}}.dy }=e^{tan^{-1}y}$

Hence required solution is

   x.I.F = $\int_{}^{} \frac{ae^{\tan^{-1}y}}{1+y^{2}}(I.F) dy$

 $\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{\tan^{-1}y}}{1+y^{2}} (e^{\tan^{-1}y}) dy$

  $\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^{2}}  dy$   ......(i)

 Put t= tan^-1 y

$\therefore$    $\frac{dt}{dy}=\frac{1}{1+y^{2}}\Rightarrow dt=\frac{1}{1+y^{2}} dy$

 $\therefore$      $\int\frac{e^{2\tan^{-1}y}}{1+y^{2}}  dy=\int e^{2t}.dt=\frac{e^{2t}}{2}+K$

 Hence equation(i)  becomes

 $x e^{\tan^{-1}y}=\frac{1}{2}e^{2t}+K$

  $\Rightarrow x e^{\tan^{-1}y}=\frac{1}{2}e^{2tan^{-1}y}+K$

 $\Rightarrow 2x e^{\tan^{-1}y}=e^{2tan^{-1}y}+K$