# General solution of differential equation : $(1+y^2)dx =(\tan^{-1}y -x)dy$

$(a)\;x= \tan ^{-1} x+1+ce^{\tan ^{-1} x} \\ (b)\;y= \tan ^{-1} x+1+ce^{\tan ^{-1} x} \\ (c)\;x= \tan ^{-1} -1+ce^{\tan ^{-1} x} \\ (d)\;y= \tan ^{-1} x-1+ce^{\tan ^{-1} x}$

## 1 Answer

$(1+y^2)\large\frac{dx}{dy}$$=\tan ^{-1} y-x (1+y^2) \large\frac{dx}{dy}$$+x=\tan ^{-1}y$
$\large\frac{dx}{dy} +\frac{x}{1+y^2}=\large\frac{\tan ^{-1}y}{1+y^2}$
$I.P= e^{\int pdy}$
$\qquad= e^{\int \Large\frac{1}{1+y}.dy}$
$\qquad= e^{\tan ^{-1}y}$
Thus the solution is $x(I.F) =\int \Large\frac{\tan ^{-1} y}{1+y^2}$ $I.F\;dy+c$
$xe^{\tan ^{-1}y} =\int \large\frac{\tan ^{-1}y}{1+y^2} e^{\tan ^{-1}y} $$dy+c \tan ^{-1}y=t \large\frac{1}{1=t^2}$$dt=dt$
$xe^{\tan ^{-1}y} =\int t.e^{t}dt+c$
$xe^{\tan ^{-1}y} = t.e^{t} -\int e^t.dt+c$
$xe^{\tan ^{-1}y} = t.e^{t}-e^t+c$
$xe^{\tan ^{-1}y} =\tan ^{-1}y e^{\tan ^{-1}y}-e^{\tan ^{-y}}+c$
$x= \tan ^{-1} -1+ce^{\tan ^{-1} x}$
Hence c is the correct answer.
answered Feb 4, 2014 by

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