Problem 92
Find the general integral of the Bernoulli equation y'-ytan…
Find the general integral of the Bernoulli equation $y'-y\tan x+y^2\cos x=0$.
Explanation of Method
Written as $y'-(\tan x)y=-(\cos x)y^2$ ($n=2$), substitute $v=1/y$.
Step 1: Substitute $v=1/y$
Dividing by $y^2$: $-v'-(\tan x)v=-\cos x$, i.e. $v'+(\tan x)v=\cos x$.
Step 2: Solve the linear equation
With $\mu(x)=\sec x$: $\dfrac{d}{dx}(v\sec x)=\sec x\cos x=1$, so $v\sec x=x+C$, i.e. $v=(x+C)\cos x$.
Result
Final answer:
\[\boxed{\,\frac{1}{y}=(x+C)\cos x,\qquad C\in\mathbb{R}}\]