Implicit differential equation

$\begin{aligned} y''e^{y'-1} & = 1 & \small \color{#3D99F6} \text{Let }y' = \frac {dv}{dx} \\ \frac {dv}{dx} e^{v-1} & = 1 \\ \int e^{v-1} dv & = \int dx \\ e^{v-1} & = x + \color{#3D99F6}C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ e^v & = ex + c_1 \\ v & = \ln (ex+ c_1) \\ \frac {dy}{dx} & = \ln (ex+ c_1) \\ y(x) & = \left(\frac {ex+c_1}e\right)\ln (ex+c_1) - x + c_2 \\ y(e) & = {\color{#3D99F6}\left(\frac {e^2+c_1}e\right)}\ln (e^2+c_1) - e + c_2 \\ & = {\color{#3D99F6}7}\ln7 + \color{#D61F06}5 & \small \color{#3D99F6} \implies \frac {e^2+c_1}e = 7 \implies c_1 = 7e-e^2 \\ & = 7 \ln(7e) -e + c_2 \\ & = 7 \ln 7 + \color{#D61F06} 7 - e + c_2 & \small \color{#D61F06} \implies 7 - e + c_2 = 5 \implies c_2 = e-2 \end{aligned}$

$\begin{aligned} \implies y(x) & = \left(x+7-e\right)\ln (e[x+7-e]) - (x +7- e) +5 \\ & = \left(x+7-e\right)\ln (x+7-e) +5 \end{aligned}$

$\begin{aligned} \implies y(e^2+e-7) & = \left(e^2+e-7+7-e\right)\ln (e^2+e-7+7-e) +5 \\ & = e^2 \ln (e^2) + 5 \\ & = 2e^2 + 5 \\ & \approx \boxed{19.7781} \end{aligned}$

$y" e^{y'-1} = 1$

This is not, obviously, a linear differential equation. Thus, we can't directly attack these using common methods which we use for second-order linear differential equations.

We can first introduce a substitution $p= y'$ . However, while it does follow that $p' = y''$ , we need to know that all differentiations in the equation are of in the independent variable, which apparently, is not in the equation. So, by chain rule,

$y" = p' = \frac{dp}{dx} = \frac{dp}{dy} \cdot \frac{dy}{dx} = p \frac{dp}{dy}$

This will simplify the equation to a first order DE, namely,

$p\frac{dp}{dy}e^{p-1} = 1$

which is now separable. This further leads us to

$pe^{(p-1)}dp = dy$

which, by taking the antiderivative of both sides, yields

$(p-1)e^{(p-1)} = y + C_1$

Now, while this may already seem to be the final solution, it isn't, since $p$ itself is a derivative of $y$ with respect to the IV (which we'll assume as $x$ ). We need to express $p$ in terms of $y$ alone.

The Lambert's W Function $W(x)$ is defined such that $x= W(xe^x)$ . That is, it is the inverse of $f(x) = xe^x$ . Taking the $W$ of both sides,

$W( (p-1)e^{(p-1)} ) = W(y+C_1)$

$p-1= W(y+ C_1)$

$p= W(y+ C_1) +1$

And now that we know $p = \frac{dy}{dx}$ , we now have

$\frac{dy}{dx} = W(y + C_1) + 1$

which is again, separable!

$\frac{dy}{W(y+ C_1) + 1} = dx$

let us introduce another substitution.

Let $y = ue^u - C_1$ .

$dy = (u+1)e^u du$

$u = W(y+C_1)$ .

This will simplify the equation further to

$\frac{(u+1)e^u du}{(u+1)} = dx$

$e^u du = dx$

$e^u = x + C_2$

which now becomes

$e^{ W(y+C_1) } = x + C_2$

$W(y+C_1) = \ln ( x + C_2 )$

Now, using the fact that $W(x)e^{W(x)} = x$ , we can change the equation to

$W(y+C_1)e^{W(y+C_1)} = \ln (x+C_2) e^{\ln (x+C_2)}$

$y+C_1 = (x+C_2) \ln (x+C_2)$

$y= (x+C_2) \ln (x+C_2) + C_1$

Note that since constants are still arbitrary at this point, there is no matter in changing the signs or not, as it will automatically follow when looking for their specific values at specific solutions.

Now, we have two initial conditions that will let us find the values of $C_1$ and $C_2$ .

When $x=e$ , we know that $y= (e+C_2) \ln (e+C_2) + C_1 = 7 \ln 7 + 5$ .

When $x=2e-7$ , we also know that $y = (2e - 7 + C_2) \ln (2e - 7 + C_2) + C_1 = e + 5$ .

Further inspection will tell us that $C_1 = 5$ , and $C_2 = 7-e$ .

Now, we have the particular solution for the differential equation above as

$y = ( x+7-e) \ln (x+7-e) + 5$

And substituting $x = e^2 + e - 7$ will give us

$y = e^2 \ln e^2 + 5 = 2e^2 + 5 = \boxed{19.7781}$