solve the given differential equation.\n5y tan x dx + 2 cos²x dy = 0\n\nchoose the correct answer…

Explanation:

Step1: Separate variables

Given $5y\tan xdx + 2\cos^{2}xdy = 0$, we can rewrite it as $\frac{5\tan x}{2\cos^{2}x}dx=-\frac{dy}{y}$. Since $\tan x=\frac{\sin x}{\cos x}$, then $\frac{5\tan x}{2\cos^{2}x}=\frac{5\sin x}{2\cos^{3}x}$.

Step2: Integrate both sides

Integrate $\int\frac{5\sin x}{2\cos^{3}x}dx=-\int\frac{dy}{y}$. Let $u = \cos x$, then $du=-\sin xdx$. So $\int\frac{5\sin x}{2\cos^{3}x}dx=-\frac{5}{2}\int u^{- 3}du$. $-\frac{5}{2}\int u^{-3}du=-\frac{5}{2}\times\frac{u^{-2}}{-2}+C_1=\frac{5}{4u^{2}}+C_1=\frac{5}{4\cos^{2}x}+C_1$. And $-\int\frac{dy}{y}=-\ln|y|+C_2$.

Step3: Simplify the result

$\frac{5}{4\cos^{2}x}+C_1=-\ln|y|+C_2$. Let $C = C_2 - C_1$, then $\frac{5}{4\cos^{2}x}+\ln|y|=C$. We know that $\frac{1}{\cos^{2}x}=\sec^{2}x$, so $\sec^{2}x+\frac{4}{5}\ln y = C$.

Answer:

A. $C=\sec^{2}x+\frac{4}{5}\ln y$