Initial value problem $y''=e^{2y}, y(0)=0, y'(0)=1$

Question

Supposedly, it is possible to determine information about the constants of this IVP solution, without computing the solution of the differential equation. Here's how I solve this.

Let $z=y'$. In brief, we then have the separable equation $$z\frac{dz}{dy}=f(y,z)=e^{2y}$$

and solving for $z$ $$z=\sqrt{e^{2y}+c}$$

which can be made into a separable equation and integrated. After playing around with the initial conditions, I did find that $c=0$. But this is a contradiction with the results of the second separable equation. Can we really solve $c$ before finding the explicit form of $y$?

Answer 1

From my answer to this, we see that we need to take the $+$ of the square root and the answer will be case $3$ where $C_1=0$. Thus the answer is

$$y=-\log(1-x)$$

Answer 2

You have $y'' = e^{2y}$. Multiplying by $y'$ you have $y'y'' = y'e^{2y}$, which you can integrate: \begin{align*} \int y'y''\, dt &= \int y' dy' \\ &=\tfrac{1}{2}(y')^2 + C \end{align*} where $C$ is a constant of integration. On the other side you have \begin{align*} \int y' e^{2y}\, dt &= \int e^{2y}\, dy \\ &= \tfrac{1}{2}e^{2y} + D \end{align*} with $D$ also a constant of integration. Combining you end up with \begin{equation*} (y')^2 = e^{2y} + 2E \end{equation*} where $E = C + D$. So if you happen to know $y(0)$ and $y'(0)$, you have your $E$.

Answer 3

By integration from $0$ to $y$, $$2zz'=2e^{2y}\implies z^2-z_0^2=e^{2y}-1$$

and

$$z=y'=e^y$$ (the solution $z=-e^y$ is not possible).

Then integrating between $0$ and $x$,

$$-e^{-y}+e^{-y_0}=x$$

or

$$y=-\log(1-x).$$