Solve this initial value problem $y'=\tan^2(5x+5y)$, with the initial condition $y(0)=0$

Question

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 $\frac{\tan(5(x+y))}{10\tan^2(5(x+y) )+10} +\frac{5x+5y}{10} -x$.

Answer 1

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.

Answer 2

$$y'=\tan^2(5x+5y)$$ 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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