Find the general solution of y'' = e^y

Find the general solution of $y'' = e^y$.

Step 1: Substitution

Let $p = y'$. Express $y''$ using the chain rule: $y'' = p\,\frac{dp}{dy}.$

The equation becomes: \[p\,\frac{dp}{dy} = e^y. \tag{1}\]

Step 2: First integration

Separate variables and integrate: \[p\,dp = e^y\,dy,\qquad \frac{p^2}{2} = e^y + C_1,\qquad p^2 = 2e^y + C,\] where $C = 2C_1$ is an arbitrary constant. Thus: \[y' = \pm\sqrt{2e^y + C}. \tag{2}\]

Step 3: Second integration — the case $C=0$

Separate variables in (2) with $C=0$: $e^{-y/2}\,dy = \pm\sqrt{2}\,dx.$

Integrate: \[-2e^{-y/2} = \pm\sqrt{2}\,x + \mathrm{const} \implies e^{-y/2} = -\frac{x+C_2}{\sqrt{2}}.\]

Square both sides: \[e^{-y} = \frac{(x+C_2)^2}{2} \implies y = \ln\frac{2}{(x+C_2)^2}.\]

Step 4: General solution

For $C = 0$:

For $C \neq 0$, the solution is given implicitly by quadrature of (2): $\pm(x + C_2) = \int \frac{dy}{\sqrt{2e^y + C}}$, where $C$ and $C_2$ are arbitrary constants (the case $C=0$ above is the elementary closed form of this family).

Step 5: Verification for $C=0$

With $y = \ln 2 - 2\ln|x+C_2|$: $y' = -\frac{2}{x+C_2},\qquad y'' = \frac{2}{(x+C_2)^2}.$

Check $y'' - e^y$: $\frac{2}{(x+C_2)^2} - \frac{2}{(x+C_2)^2} = 0.$ $\checkmark$