Answer
$y=\ln(\ln (\sec x + \tan x)+\frac{1}{C}\ln (\sec x + \tan x)+C_2$
Work Step by Step
We are given:
$y′′+y'\tan x=(y')^2$
Suppose that
$a=\frac{dy}{dx}$
Now, $\frac{da}{dx}=\frac{d^2y}{dx^2}=\frac{da}{dy}.\frac{da}{dx}=a\frac{da}{dy}$
Substitute into the equation:
$\frac{da}{dx}+a\tan x=a^2$
Dividing both sides by $a^2$
$\frac{1}{a^2}\frac{da}{dx}+\frac{1}{a}\tan x=1$
Let $t=\frac{1}{a} \rightarrow \frac{dt}{dx}=-\frac{1}{a^2}\frac{da}{dx}$
The equation becomes:
$ \frac{dt}{dx} -t \tan x=-1$
Integrating factor:
$I(x)= e^{\tan x dx }=e^{\ln |\sec x|}=\sec x$
Hence,
$\frac{d}{dx}(t \sec x)=-\sec x$
Integrating both sides:
$t \sec x = \ln (\sec x +\tan x)+C$
$t=\frac{ \ln (\sec x +\tan x)+C}{\sec x}$
$a=\frac{\sec x }{\ln |\sec x +\tan x|}+\frac{\sec x}{C}$
$\frac{dy}{dx}=\frac{\sec x}{\ln |\sec x + \tan x|}+\frac{\sec x}{C}$
Continue to integrate:
$y=\int \frac{\sec x}{\ln |\sec x +\tan x|}dx+\int \frac{\sec x}{C}dx$
$y=\int \frac{1}{t}dt+\frac{1}{C}\int \sec x dx$
The final solution is:
$y=\ln(\ln (\sec x + \tan x)+\frac{1}{C}\ln (\sec x + \tan x)+C_2$
