Chaya Galloway

2021-03-07

Solve differential equation $\frac{{\mathrm{cos}}^{2}y}{4x+2}dy=\frac{\left(\mathrm{cos}y+\mathrm{sin}y{\right)}^{2}}{\sqrt{{x}^{2}+x+3}}dx$

mhalmantus

$\frac{{\mathrm{cos}}^{2}y}{4x+2}dy=\frac{{\mathrm{cos}}^{2}y+{\mathrm{sin}}^{2}y+2\mathrm{sin}y\mathrm{cos}y}{\sqrt{{x}^{2}+x+3}}dx$
$\frac{{\mathrm{cos}}^{2}y}{4x+2}dy=\frac{1+\mathrm{sin}2y}{\sqrt{{x}^{2}+x+3}}dx$
$\frac{\mathrm{cos}2y}{1+\mathrm{sin}2y}dy=\frac{4x+2}{\sqrt{{x}^{2}+x+3}}dx$
Integrating on both sides
$\int \frac{\mathrm{cos}2y}{1+\mathrm{sin}2y}dy=\int \frac{4x+2}{\sqrt{x2+x+3}}dx$
$\int \frac{4x+2}{\sqrt{x2+x+3}}dx$

$2x+1dx=2udu$
$4x+2dx=4udu$

$=4du=4u+c$
$=4\sqrt{{x}^{2}+x+3}+c$
$\int \frac{\mathrm{cos}2y}{1+\mathrm{sin}2y}y$

Differentiating
$dt=2\mathrm{cos}2y$
$\int \frac{\mathrm{cos}2y}{1+\mathrm{sin}2y}dy=\int \frac{\frac{1}{2}dt}{t}$
$=\frac{1}{2}\int \frac{1}{t}dt$ $=\frac{1}{2}\mathrm{ln}t+K$
$\frac{1}{2}\mathrm{ln}\left(1+\mathrm{sin}2y\right)+k$
$\int \frac{\mathrm{cos}2y}{1+\mathrm{sin}2y}dy=\int \frac{4x+2}{\sqrt{x2+x+3}}dx$
$4\sqrt{x2+x+3}=\frac{1}{2}\mathrm{ln}\left(1+\mathrm{sin}2y\right)+c$

Jeffrey Jordon