If $y = e^{5x} + e^{-5y}$, prove that $y'' = 25y$

Question

Question:
$y = e^{5x} + e^{-5y}$, prove that $y'' = 25y$

what i have done:
$y - e^{-5y}= e^{5x} $ $$(1 + 5e^{-5y})y'= 5e^{5x} $$ $$y'= \frac{5e^{5x}}{(1 + 5e^{-5y})} $$ $$y'' = 5\frac{\bigg[ 5e^{5x}(1+5e^{-5y}) - e^{5x}(-25e^{-5y})y' \bigg]}{(1+5e^{-5y})^2}$$ $$y'' = 25\frac{\bigg[ e^{5x}(1+5e^{-5y}) + e^{5x}(5e^{-5y})\frac{5e^{5x}}{(1 + 5e^{-5y})}) \bigg]}{(1+5e^{-5y})^2}$$

$$y'' = 25{\bigg[ e^{5x}\frac{1}{(1+5e^{-5y})} + e^{-5y}\frac{25e^{5x}e^{5x}}{(1+5e^{-5y})^3} \bigg]}$$

This last step here tells me I have gone wrong somewhere, somehow. But I just can't figure out where.
Any hints to get the correct answer??

Answer 1

The statement is incorrect. If $y''=25y$ then the solutions are $y=Ae^{5x}+Be^{-5x}$ for constants $A,B$. This means that \begin{align}y=e^{5x}+e^{-5y}&\implies Ae^{5x}+Be^{-5x}=e^{5x}+e^{-5Ae^{5x}+5Be^{-5x}}\quad\forall x\\&\implies-5Ae^{5x}+5Be^{-5x}=\ln((A-1)e^{5x}+Be^{-5x})\quad\forall x\end{align} which is not true.