Solve this initial value problem $y'=\tan^2(5x+5y)$, with the initial condition $y(0)=0$ 
So I'm about to lose my hair, I already have 60 attempts done for this problem. 
I took $u= 5x+5y$, which gave me $u'= 5+5y'$.
Then I isolated $y'= (u'-5)/5$, which would give me 
$5\tan^2(u)+5 = u'$ and then I have integral of $dx = \int \frac{du}{5\tan^2(u)+5}$.
And after by using an integral calculator I get that $x +C= \frac{\tan(u)}{10\tan^2(u) +10} +\frac u{10}$. 
By replacing the $u$ back and using the initial condition $y(0)=0$, I get $C=0$ and the solution that I found was $\frac{\tan(5(x+y))}{10\tan^2(5(x+y) )+10} +\frac{5x+5y}{10} -x$.
 A: Let us observe that
\begin{align}
y' =\tan^2(5x+5y) \ \ \implies&  \ \ \cos^2(5x+5y)y' = \sin^2(5x+5y) = 1-\cos^2(5x+5y)\\
\implies& \ \ \cos^2(5x+5y)(1+y') = 1.
\end{align}
Next, notice if we define $u(x) =5y(x)+5x$, then it follows
\begin{align}
\cos^2(5x+5y)(1+y') = 1 \ \ \implies \ \ \cos^2(u)u' = 5 \ \ \implies \ \ \frac{1+\cos(2u)}{2} du = 5 dx.
\end{align}
Hence the general implicit solution is given by
\begin{align}
\frac{1}{2}u + \frac{1}{4}\sin(2u)= 5x+C \ \ \implies \ \ \frac{5}{2}(y+x) + \frac{1}{4}\sin(10(x+y))= 5x+C.
\end{align}
Plugging in the initial condition yields
\begin{align}
F(x, y) = \frac{5}{2}(y-x) + \frac{1}{4}\sin(10(x+y))+1 = 1.
\end{align}
Edit: The problem looks ill-posed since
\begin{align}
F(x, y) = \frac{1}{2}(y-x) + \frac{1}{20}\sin(10(x+y))+1 = 1.
\end{align}
also works. 
A: $$y'=\tan^2(5x+5y)$$
Let $u(x)=5x+5y(x)\quad\implies\quad y(x)=\frac15 u(x)-x$
$$y'=\frac15 u'-1=\tan^2(u)\quad\implies\quad \frac15 u'=\frac{1}{\cos^2(u)}$$
This is a separable ODE : $\quad \cos^2(u)du=5dx$
$ \int\cos^2(u)du=\frac14(2u+\sin(2u))=5x+\text{constant} $
$2(5x+5y)+\sin(2(5x+5y))-20x=c$
This is the solution on the form of implicit equation :
$$10(y-x)+\sin(10(x+y))=c$$
The solution cannot be expressed on explicit form $y(x)$ in terms of a finite number of standard functions.
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