# Classify the following differential equations as separable, homogeneous, parallel line,

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. $2x{\mathrm{sin}}^{2}ydx-\left({x}^{2}-9\right)\mathrm{cos}ydy=0$
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Philip Williams
Given differential equation
$2x{\mathrm{sin}}^{2}ydx-\left({x}^{2}-9\right)\mathrm{cos}ydy=0$
write the equation in the standard form
$\frac{dy}{dx}=\frac{2x{\mathrm{sin}}^{2}y}{\left({x}^{2}-9\right)\mathrm{cos}y}$
we can saperate the variable so, equation is seperable now seperating the variable
$\left(\frac{\mathrm{cos}y}{{\mathrm{sin}}^{2}y}\right)dy=\left(\frac{2x}{{x}^{2}-9}\right)dx$
integrate both side
$⇒\int \left(\frac{\mathrm{cos}y}{{\mathrm{sin}}^{2}y}\right)dy=\int \left(\frac{2x}{{x}^{2}-9}\right)dx$
$⇒\int \frac{d\left(\mathrm{sin}y\right)}{{\mathrm{sin}}^{2}y}=\int \frac{d\left({x}^{2}-9\right)}{{x}^{2}-9}$
$⇒-\frac{1}{\mathrm{sin}y}=\mathrm{ln}\left({x}^{2}-9\right)+C$

hence solution of differential equation is
###### Not exactly what you’re looking for?
temnimam2
$2x{\mathrm{sin}}^{2}ydx-\left({x}^{2}-9\right)\mathrm{cos}ydy=0$
$\therefore 2x{\mathrm{sin}}^{2}ydx=\left({x}^{2}-9\right)\mathrm{cos}ydy$
$\therefore \frac{2x}{\left({x}^{2}-9\right)}dx=\frac{\mathrm{cos}y}{{\mathrm{sin}}^{2}y}dy$

This is variable separable form.
$\therefore$ the general solution is
$\int \frac{2x}{{x}^{2}-9}dx=\int \mathrm{cos}ecy\cdot \mathrm{cot}ydy$

This is general solution.
karton