What is the value of $x$ such that $\frac{\text{d}^2y}{\text{d}x^2}=0$ where $\frac{\text{d}y}{\text{d}x}=-ae^{-bx}y-cy+d$? How can you find the values of $x$ such that
$$\frac{\text{d}^2y(x)}{\text{d}x^2}=0$$
where
$$\frac{\text{d}y}{\text{d}x}=-ae^{-bx}y-cy+d$$
with
$$y(0)=y_0$$
and
$$a,b,c,d>0$$
If it helps I can approximate $b\approx c$, I am currently working on an solution that utilises that fact but it seems to have failed. I will post any advances I make but currently I am stumped.
The DE isn't particularly nicely solvable as you end up with exponential integrals. This problem is also  equivalent to maximising the first derivative of $y$, as I know this equation produces a sigmoid shape. I also would not mind an approximation for the solutions of $x$.
My workings:
If we define
$$f(x)=ae^{-bx}+c$$
and
$$g(x)=e^{\int f(x) \Bbb dx}$$
then we find
$$\frac{\text{d}y}{\text{d}x}=d\frac{\text{d}}{\text{d}x}
\left(
\frac{d}{g(x)}
\int_0^x g(s)\text{d}s
\right)
$$
Then by differentiating once and setting it equal to zero we find the problem reduces to solving
$$g^2g'+gg''\int_0^x g(s) \Bbb d s=2g'^2\int_0^x g(s) \Bbb d s$$
where primes indicate differentiation.
 A: $$\frac{dy}{dx}=-ae^{-bx}y-cy+d$$
Multiplying both sides by $e^{bx}$ and rewrite the D.E to be in the form
$$(ay+(cy-d)e^{bx})dx+e^{bx}dy=0$$
Let $M(x,y)=ay+(cy-d)e^{bx} \Rightarrow M_y=a+ce^{bx}$, and
$N(x,y)=e^{bx} \Rightarrow N_x=be^{bx}$, which mean that the D.E. is not exact.
Let find an appropriate integrating factor $\mu(x)$ (depends only on $x$) for the which the D.E. is exact
$$\frac{d\mu}{\mu}=\frac{M_y-N_x}{N}=\frac{a+ce^{bx}-be^{bx}}{e^{bx}}=ae^{-bx}+c-b$$
Thus, $$\ln \mu(x)= -\frac{a}{b}e^{-bx}+(c-b)x\Rightarrow\mu(x) =e^{-\frac{a}{b}e^{-bx}}\cdot e^{(c-b)x}$$
Multiplying both sides of (1), by $\mu(x)$ we get an exact D.E.
So that, let $\phi(x,y)$ be a solution of the D.E. such that 
$\frac{\partial \phi}{\partial y}=N(x,y)=e^{-\frac{a}{b}e^{-bx}}\cdot e^{(c-b)x} \cdot e^{bx}=e^{-\frac{a}{b}e^{-bx}}\cdot e^{cx} \Rightarrow \phi(x,y)=e^{-\frac{a}{b}e^{-bx}}\cdot e^{cx}y+h(x)$
ut since $\phi_x=M(x,y)$, we get that $$
e^{ - {\textstyle{a \over b}}e^{ - bx} } e^{cx} \left( {c + ae^{ - bx} } \right)
y+h'(x)= 
aye^{ - {\textstyle{a \over b}}e^{ - bx} } e^{\left( {c - b} \right)x}  + \left( {cy - d} \right)e^{ - {\textstyle{a \over b}}e^{ - bx} } e^{cx} 
$$
$$\Rightarrow h'(x)=-d
e^{ - {\textstyle{a \over b}}e^{ - bx} } e^{cx} 
\Rightarrow  h(x)=-d\int{e^{ - {\textstyle{a \over b}}e^{ - bx} } e^{cx}  dx}$$
Hence, $$\phi(x,y)=e^{-\frac{a}{b}e^{-bx}}\cdot e^{cx}y-d\int{e^{ - {\textstyle{a \over b}}e^{ - bx} } e^{cx}  dx}=C$$ where, $C$ is constant
A: There is no need to integrate or use boundary conditions thereby introducing new constants.  Just differentiate once, set second derivative to zero, algebraically simplify it to get an implicit relation between $x , y$ as follows:
$$ \frac{dy}{dx}=-ae^{-bx}y-cy+d \tag{1} $$
Let $ e^{b\,x}= u $ so that $ u{'}= b\, u.$ Primes denote differentiation with respect to x. $ a,b,c,d $ are constants. $ y= y _{PI} $ for short.
$ y^{''}$ vanishes when  $ y^{'}$ is an extremum. Also if  $ y^{'}$ is in a quotient form, then  using Quotient Rule,
$$ y^{'}= (\dfrac pq)^{'} =  \dfrac {q p ^{'}-q^{'} p  }{q^2} = 0\rightarrow \frac pq =  \frac {p^{'}}{q^{'}} $$
$$ \dfrac{-a y -cyu+du}{u} =\dfrac{-ay^{'}- c ( y^{'}u + y u^2 b) + dub}{ub}  \tag{2} $$
Removing negative sign
$$ \dfrac{a y + cyu-du}{u} =\dfrac{ay^{'}+ c ( y^{'} u + y u^2 b) -dub }{ub}  $$
$$ aby - bcyu = a y^{'} + c u ( y^{'} + buy) \tag{3}$$
$$ by (a- c u^2 -cu )= y^{'} (a+ c u) $$
Plugging in from (1)
$$ by (a- c u^2 -cu )= ( -a y/u-c y+d ) (a+ c u) \tag{4}$$
is the required implicit function  $ f(x,y_{PI})= 0 $ for inflection point loci which may include them in the sigmoid plot. To find specific $x$ values cubic equations have to be  solved further.
EDIT 1:
For a special case taking values $ a = 1, c = d =0,$  all $x_{PI}$ are on a line parallel to y-axis at a distance of $$ x_{PI}= -\frac{log (b)}{b}. $$
From mwomath's expression
$$ y = C Exp[ \frac{-e^{-b x}}{b}] $$
EDIT 1:
Giving a simpler example to identify inflection locus of a given DE:
$$ y' = - x\, y $$
Integrating, you get
$$ y = C e^{\frac{-x^2}{2}} $$
which is a Gaussian probability curve of variable maximum height $C.$
Take next (second order) derivative and eliminate $y'$ to get:
$$ y '' = y ( 1- x^2) $$
for inflection locus, $ y^{''}=0. $
Plotting its graph of the curve with several values for $C$ you can see two neat lines parallel to y-axis $$ x = \pm 1, $$ each passing through the inflection point of each corresponding curve on either side.
