# Solve differential equationdx/dy+x/y= 1/(sqrt(1+y^2))

Solve differential equation $\frac{dx}{dy}+\frac{x}{y}=\frac{1}{\sqrt{1+{y}^{2}}}$

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$\frac{dx}{dy}+p\left(y\right)x=q\left(y\right)$
Compare the equation $\frac{dx}{dy}+p\left(y\right)x=q\left(y\right)$ to $\frac{dx}{dy}+\frac{x}{y}=\frac{1}{\sqrt{1+{y}^{2}}}$ and obtain $p\left(y\right)=\frac{1}{y}$, $q\left(y\right)=\frac{1}{\sqrt{1+{y}^{2}}}$
$I.F.={e}^{\int p\left(y\right)dy}$
$={e}^{\mathrm{ln}y}$
=y $x{e}^{\int p\left(y\right)dy}=\int {e}^{\int p\left(y\right)dy}q\left(y\right)dy+C$
where C is arbitrary constant of equation
$x{e}^{\int p\left(y\right)dy}=\int {e}^{\int p\left(y\right)dy}q\left(y\right)dy+C$
$xy=\int y\frac{1}{\sqrt{1+{y}^{2}}}dy+C$
$xy=\frac{1}{2}\int \frac{2y}{\sqrt{1+{y}^{2}}}dy+C$
$xy=\frac{1}{2}\int \frac{\left(1+{y}^{2}{\right)}^{\prime }}{\sqrt{1+{y}^{2}}}dy+C$ [$\because \left(1+{y}^{2}{\right)}^{\prime }=2y$]
$xy=\frac{1}{2}\left(2\sqrt{1+{y}^{2}}\right)+C$ [$\because \int \frac{f\left(y\right)}{\sqrt{f\left(y\right)}}dy=2\sqrt{f\left(y\right)}$]
$xy=\sqrt{1+{y}^{2}}+C$
$x=\frac{\sqrt{1+{y}^{2}}+C}{y}$
$x=\frac{\sqrt{1+{y}^{2}}}{y}+\frac{C}{y}$