# Solve for the differential equations and get the general solution.

Solve for the differential equations and get the general solution. Simplify your answer free from In. $dy=\mathrm{tan}x\mathrm{tan}ydx$
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vrangett

$\frac{dy}{\mathrm{tan}y}=\mathrm{tan}xdx$
$\frac{dy}{\frac{\mathrm{sin}y}{\mathrm{cos}y}}=\mathrm{tan}xdx$

$\mathrm{ln}\left(\mathrm{sin}y\right)=-\mathrm{ln}\left(\mathrm{cos}x\right)+C$
${e}^{\mathrm{ln}\left(\mathrm{sin}y\right)}={e}^{-\mathrm{ln}\left(\mathrm{cos}x\right)+C}$
$\mathrm{sin}y=\frac{{e}^{{\mathrm{ln}\left(\mathrm{cos}x\right)}^{-1}}}{{e}^{C}}$
$\mathrm{sin}y={\left(\mathrm{cos}x\right)}^{-1}.{e}^{C}$
$\mathrm{sin}y=\frac{{e}^{C}}{\mathrm{cos}x}$
$y={\mathrm{sin}}^{-1}\left(\frac{{e}^{C}}{\mathrm{cos}x}\right)$
Answer: The general solution is $y={\mathrm{sin}}^{-1}\left(\frac{{e}^{C}}{\mathrm{cos}x}\right)$
###### Not exactly what you’re looking for?
Hattie Schaeffer
Given differential equation

$⇒\frac{\mathrm{cos}y}{\mathrm{sin}y}dy=\frac{\mathrm{sin}x}{\mathrm{cos}x}dx$
Integrating, $\int \frac{\mathrm{cos}y}{\mathrm{sin}y}dy=\int \frac{\mathrm{sin}x}{\mathrm{cos}x}dx$ (1)
We substitute left hand integral
$u=\mathrm{sin}y$

and right hand integral
$\mathrm{cos}x=v$

Then from (1)
$\int \frac{du}{u}=\int \frac{-dv}{v}$
$⇒\mathrm{ln}\left(u\right)=-\mathrm{ln}\left(v\right)+\mathrm{ln}\left(c\right)$
$⇒\mathrm{ln}\left(u\right)=\mathrm{ln}\left(\frac{c}{v}\right)$
$⇒u=\frac{c}{v}$
$⇒vu=c$
$⇒\mathrm{cos}x\mathrm{sin}y=c$
Hence solution is $\mathrm{cos}x\mathrm{sin}y=c$
karton