# Find the general solution of the differential equation $(1+y^2)+(x-e^{\large\tan^{-1}y})\Large \frac{dy}{dx}$=0

Toolbox:
• A linear differential equation of the form $\large\frac{dx}{dy}$$+Py=Q, has the general solution as xe^{\int pdy}=\int Q. e ^{\int pdy}.dy+c Where e^{\int pdy} is the integrating factor (I.F) • \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c$
Step 1:
Given $(1+y^2)+(x-e^{\tan ^{-1}y})\large\frac{dy}{dx}=0$
This can be written as
$\large\frac{dx}{dy}=\frac{(x-e^{\Large\tan ^{-1}y})}{1+y^2}$
$\large\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\Large\tan ^{-1}y}}{1+y^2}$
Clearly this represents a linear differential equation which is of the form
$\large\frac{dx}{dy}$$+Px=Q Here P= \large\frac{+1}{1+y^2} and Q=\large\frac{e^{\Large\tan ^{-1}y}}{1+y^2} Let us first find the integrating factor I.F which e^{\int pdy} e^{\large\int pdy}=e^{\Large\int \frac{1}{1+y^2}dy} Consider \int \large\frac{1}{1+y^2}$$dy$
On integrating this we get $\tan ^{-1}y$
Therefore $e^{\large \int pdy}=e^{\large\tan ^{-1}y}$
The solution for the linear differential equation is
Step 2:
$x.e^{\large\int pdy}=\int Q.e^{\large\int pdy}dy+c$
Now substituting the I.F we get
$x.e^{\large\tan ^{-1}y}=\int \large\frac{e^{\Large\tan ^{-1}y}}{1+y^2}.e^{\large\tan ^{-1}y}$$dy+c =\int \large\frac{e^{\Large2\tan ^{-1}y}}{1+y^2}$$dy+c$
Put $\tan ^{-1}y=t,$ on differentiating w.r.t x we get,
$\large\frac{dy}{1+y^2}$$=dt Now substituting this we get, x.e^{\large\tan ^{-1}y}=\int e^{2t}.dt On integrating we get x.e^{\large\tan ^{-1}y}=\large\frac{1}{2}\int e^{2t}$$+c$
=>$2x.e^{\large\tan ^{-1}y}= e^{2t}$$+c Substituting for t we get, =>2x.e^{\large\tan ^{-1}y}= e^{\large2\tan ^{-1}y}$$+c$
Hence the required solution is
$2x.\tan ^{-1}y= e^{\large2\tan ^{-1}y}$$+c$