Inyalan0

2021-12-30

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. $2x{\mathrm{sin}}^{2}ydx-\left({x}^{2}-9\right)\mathrm{cos}ydy=0$

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

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