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Solve the following.$ \large\frac{dx}{dy}=\frac{x}{1+y^2}=\frac{\tan^{-1}}{1+y^2}$

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  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1
This is a linear DE in $\large\frac{dx}{dy}$
where $P(y) =\large\frac{1}{(1+y^2 )}$
Step 2:
$\therefore \int \large\frac{1}{1+y^2 }$$dy=\tan^{-1} y$
Step 3:
The solution is $ x \tan ^{-1}y=\int e^{\tan ^{-1}y} \large\frac{\tan^{-1}y}{1+y^2}$$dy$
$\qquad=\int {\tan ^{-1}y} \large\frac{e^{\Large \tan^{-1}y}}{1+y^2}$$dy$
$\qquad =\tan ^{-1}y e^{\tan ^{-1}y} -\int \large\frac{e^{\Large\tan ^{-1}y}}{1+y^2}$$dy$
$\qquad =\tan ^{-1}y e^{\tan ^{-1}y} -e^{\large\tan ^{-1}y}$$+c$
The GS is $x \tan^{-1} y=\tan^{-1} y (e^{\large\tan^{-1} y}-1) + c$
answered Sep 5, 2013 by meena.p
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