Finding analytic expression for $y'(x)$ from $y''(x) = \exp(a\cdot y(x))$

Question

Is it possible to integrate $$y''(x) = \exp(a y(x)) \ \ \ \ \tag{1}$$ to get $$ y'(x) = \frac{\exp(ay(x))}{ay'(x)} + C\ ? \ \ \ \ \tag{2}$$

And then solve the equation $$ y'(x)^2-Cy'(x) = \frac{\exp(ay(x))}{a} \ \ \ \ \tag{3}$$ to get $$ y'(x) = \frac C2\pm \sqrt{\frac{C^2}{4}+\frac{\exp(ay(x))}{a}}\ ? \ \ \ \ \tag{4}$$

I feel that the integration step might be invalid since taking the derivative of (2) (and try to return to (1)) creates some problem due to the inner derivative of the denominator: $$ y''(x) = \frac{\mathrm{d}\left(\frac{\exp(ay(x))}{ay'(x)}\right)}{\mathrm{d}x} = \frac{ay'(x)\exp(ay(x))}{ay'(x)}-\frac{y''(x)\exp(ay(x))}{ay'(x)^2} \ \ \ \ \tag{5}$$ $$ y''(x)\left(1+\frac{\exp(ay(x))}{ay'(x)^2}\right) = \exp(ay(x)) \ \ \ \ \tag{6}$$ $$ y''(x) = \frac{ay'(x)^2\exp(ay(x))}{ay'(x)^2+\exp(ay(x)))} \ \ \ \ \tag{7}$$

There is clearly an error somewhere as (7) does not seem to be equal to (1), can you help me spot it?

Answer 1

Your step from $(1)$ to $(2)$ is wrong !

From $y''(x) = \exp(ay(x))$

we get

$2a y'(x)y''(x)=2a y'(x)\exp(ay(x))$.

Hence

$(ay'(x)^2)'=2(\exp(ay(x)))'$

thus

$ay'(x)^2=2\exp(ay(x))+C$.

Answer 2

No this is wrong since $y$ is dependent to $x$. The right way is as following $$2y''y'=2y'e^{ay}$$by integrating$$(y')^2=2e^{ay}+C$$then according to initial conditions we have$$y'=\pm\sqrt{2e^{ay}+C}$$or $${y'\over{\sqrt {2e^{ay}+C}}}=\pm1$$using WolframAlpha we have $$ {1\over a\sqrt C}\ln {\sqrt C-\sqrt {2e^{ay}+C}\over\sqrt C+\sqrt {2e^{ay}+C}}=\pm t+C_1$$