General
Question: How to solve $a \cdot x^{2} + b \cdot x + c + \exp\left( d \cdot x + e \right) = 0$ for $x$ where $x \in \mathbb{C} \cup \left\{ \hat{\infty} \right\} \wedge \left\{ a,\, b,\, c,\, d,\, e \right\} \in \mathbb{C}$ (note: here $e$ is not euler's constant)?
Backgrund: When I sometimes calculate with ODEs, I sometimes come across equations of this form. So I'm wondering what a general solution to this would look like?
My Trys
Since I've often encountered such equations, I've tried a few things accordingly. I'll just name the best (for non-trivial stuff):
If $a = 0$
$$
\begin{align*}
0 \cdot x^{2} + b \cdot x + c + \exp\left( d \cdot x + e \right) &= 0\\
b \cdot x + c + \exp\left( d \cdot x + e \right) &= 0\\
\exp\left( d \cdot x + e \right) &= -c – x \cdot b\\
\end{align*}
$$
$$
\begin{align*}
-x \cdot b – c &= \exp\left( e – \frac{c \cdot d}{b} + x \cdot d + \frac{c \cdot d}{b} \right)\\
-x \cdot b – c &= \exp\left( e – \frac{c \cdot d}{b} + x \cdot d + \frac{c \cdot d}{b} \right)\\
-x \cdot b – c &= \exp\left( e – \frac{c \cdot d}{b} \right) \cdot \exp\left( x \cdot d + \frac{c \cdot d}{b} \right)\\
\end{align*}
$$
$$
\begin{align*}
-x \cdot b \cdot \exp\left( -x \cdot d – \frac{c \cdot d}{b} \right) – c \cdot \exp\left( -x \cdot d – \frac{c \cdot d}{b} \right) &= \exp\left( e – \frac{c \cdot d}{b} \right)\\
-x \cdot d \cdot \exp\left( -x \cdot d – \frac{c \cdot d}{b} \right) – \frac{c \cdot d}{b} \cdot \exp\left( -x \cdot d – \frac{c \cdot d}{b} \right) &= \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right)\\
\left( -x \cdot d – \frac{c \cdot d}{b} \right) \cdot \exp\left( -x \cdot d – \frac{c \cdot d}{b} \right) &= \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right)\\
\end{align*}
$$
$$
\begin{align*}
-x_{k} \cdot d – \frac{c \cdot d}{b} &= W_{k}\left( \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right) \right)\\
x_{k} + \frac{c}{b} &= -\frac{1}{d} \cdot W_{k}\left( \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right) \right)\\
x_{k} &= -\frac{1}{d} \cdot W_{k}\left( \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right) \right) – \frac{c}{b}\\
\end{align*}
$$
$$\fbox{$x_{k} = -\frac{1}{d} \cdot W_{k}\left( \frac{d}{b} \cdot \exp\left( e – \frac{c \cdot d}{b} \right) \right) – \frac{c}{b}$}$$
Where $W_{k}$ is the Lambert W Function…
If $a \ne 0$
Completing …
$$
\begin{align*}
a \cdot x^{2} + b \cdot x + c + \exp\left( d \cdot x + e \right) &= 0\\
a \cdot x^{2} + b \cdot x + \exp\left( d \cdot x + e \right) &= -c\\
a \cdot x^{2} + b \cdot x + \frac{b^{2}}{4} + \exp\left( d \cdot x + e \right) &= -c + \frac{b^{2}}{4}\\
a \cdot x^{2} + b \cdot x + \frac{b^{2}}{4} + \exp\left( d \cdot x + e \right) &= -c + \frac{b^{2}}{4}\\
x^{2} + \frac{b}{a} \cdot x + \frac{b^{2}}{4 \cdot a^{2}} + \frac{1}{a} \cdot \exp\left( d \cdot x + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}}\\
\left(x + \frac{b}{2 \cdot a} \right)^{2} + \frac{1}{a} \cdot \exp\left( d \cdot x + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}}\\
\end{align*}
$$
Fail: I can't think of anything useful to add.
Substitution + Compliting …
$$
\begin{align*}
\left(\underbrace{x + \frac{b}{2 \cdot a}}_{= u} \right)^{2} + \frac{1}{a} \cdot \exp\left( d \cdot x + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}}\\
u^{2} + \frac{1}{a} \cdot \exp\left( d \cdot \left( u – \frac{b}{2 \cdot a} \right) + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}}\\
u^{2} + \frac{1}{a} \cdot \exp\left( d \cdot u – \frac{d \cdot b}{2 \cdot a} + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}}\\
\frac{1}{a} \cdot \exp\left( d \cdot u – \frac{d \cdot b}{2 \cdot a} + e \right) &= -\frac{c}{a} + \frac{b^{2}}{4 \cdot a^{2}} – u^{2}\\
\exp\left( d \cdot u – \frac{d \cdot b}{2 \cdot a} + e \right) &= -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2}\\
\exp\left( d \cdot u + f – \frac{d \cdot b}{2 \cdot a} + e – f \right) &= -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2}\\
\exp\left( d \cdot u + f \right) \cdot \exp\left( -\frac{d \cdot b}{2 \cdot a} + e – f \right) &= -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2}\\
\exp\left( -\frac{d \cdot b}{2 \cdot a} + e – f \right) &= \left( -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2} \right) \cdot \exp\left( -d \cdot u – f \right)\\
\end{align*}
$$
Fail: Nothing to simplify…
Despair
$$
\begin{align*}
\exp\left( -\frac{d \cdot b}{2 \cdot a} + e – f \right) &= \left( -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2} \right) \cdot \exp\left( -d \cdot u – f \right)\\
-\frac{d \cdot b}{2 \cdot a} + e – f &= \ln\left( \left( -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2} \right) \cdot \exp\left( -d \cdot u – f \right) \right)\\
-\frac{d \cdot b}{2 \cdot a} + e – f &= \ln\left( -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2} \right) + \ln\left( \exp\left( -d \cdot u – f \right) \right)\\
-\frac{d \cdot b}{2 \cdot a} + e – f &= \ln\left( -c + \frac{b^{2}}{4 \cdot a} – a \cdot u^{2} \right) – d \cdot u – f \\
\end{align*}
$$
So there might be a solution in terms of Wright Lambert W functions…
Best Answer
Single Sum:
We solve: $$a x^2 + b x + c + \exp(d x + e)=0\iff \exp(-dx)\left(x^2+\frac ba x+\frac ca\right)=-\frac{\exp(e)}a$$
From this point on, $e$ is euler’s constant:
$$e^{-dx}(x+p)(x+q)=r\mathop\iff^{x=t-p} e^{-d(t-p)}(t-p+p)(t-p+q)=e^{-dt+dp} t(t-p+q)=e^{dp}e^{-dt} t(t-p+q)=r$$
Multiply both sides by $e^{-dp}$ and use Lagrange reversion:
$$e^{-dt}t(t-p+q)=re^{-dp}\iff t=\frac{r e^{dt}}{e^{dp}(t+q-p)}=\frac{u e^{dt}}{(t+v)}=\sum_{n=1}^\infty\frac{u^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\left(\frac{e^{dw}}{w+v}\right)\right|_0$$
Using general Leibniz rule:
$$\left.\frac{d^{n-1}}{dw^{n-1}}\left(\frac{e^{dw}}{w+v}\right)\right|_0=\sum_{k=0}^n\binom{n-1}k\left.\frac{d^{n-1-k}e^{dnw}}{w^{n-1-k}}\right|_0\left.\frac{d^k(x+v)^{-n}}{dw^k}\right|_0$$
and sum to get Bessel $\operatorname K_v(z)$. Therefore:
$$\boxed{e^{-d t}t(t+v)=u\implies t=\frac1{\sqrt\pi}\sum_{n=1}^\infty\frac{\sqrt{-dvn}(dn)^{n-1}}{n!e^\frac{dvn}2}\left(\frac uv\right)^n\operatorname K_{\frac12-n}\left(-\frac{dvn}2\right)}$$
shown here
$j$th Complex Solution:
$$a x^2 + b x + c + \exp(d x + e)=0\iff \exp(-dx)\left(x^2+\frac ba x+\frac ca\right)=\exp(-d x)(x^2+A x+B)=C$$
From this point on, $e$ is euler’s constant. Use an $x=t-\frac A2$ Tshirnhaus transformation:
$$e^{-d t+\frac{Ad}2}\left(t^2-\frac{A^2}4+B\right)=C\iff e^{-d t} \left(t^2-\frac{A^2}4+B\right)=C e^{-\frac{Ad}2}$$
Solve for $t$ and use the complex logarithm where $j\in\Bbb Z$:
$$t=-\frac1d\ln\left(\frac{e^{-\frac{Ad}2}C}{t^2-\frac{A^2}4+B}\right)-\frac{2\pi i j}d=r \ln\left(\frac p{t^2-q}\right)+2\pi i jr=2\pi i j r+\sum_{n=1}^\infty \frac{r^n}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}\ln^n\left(\frac p{t^2-q}\right)\right|_{2\pi i j r}$$
Using Stirling S1 and binomial theorem:
$$\frac1{n!}\frac{d^{n-1}}{dt^{n-1}}\ln^n\left(\frac p{t^2-q}\right)\mathop=^{\left|\frac p{t^2-q}-1\right|<1}\sum_{k=n}^\infty\frac{S_k^{(n)}}{k!}\frac{d^{n-1}}{dt^{n-1}}\left(\frac p{t^2-q}-1\right)^k= \sum_{k=n}^\infty \sum_{m=0}^k\frac{S_k^{(n)}}{k!}(-1)^{k-m} p^m\binom k m \frac{d^{n-1}}{dt^{n-1}}((t-\sqrt q)(t+\sqrt q))^{-m}$$
shown here. We evaluate $\frac{d^{n-1}}{dt^{n-1}}(t^2-q)^{-m} $ with the Gauss hypergeometric function, expressible with Legendre P for very specific values. Finally, use $\sum\limits_{n=1}^\infty \sum \limits_{k=n}^\infty a_{k,n}=\sum_ \limits{k=1}^\infty\sum \limits_{n=0}^ka_{k,n}$:
$$\boxed{e^{-d t}(t^2-q)=p\implies t_j=\frac{2\pi i j}d+\sum_{k=1}^\infty \sum_{n=0}^k\sum_{m=0}^k\frac{S_k^{(n)}\Gamma(m+n-1)(-1)^{n+k+1}p^m r^n}{(k-m)!\Gamma(m)m!\left(\frac{2\pi i j}d+\sqrt q\right)^{n-1}\left(\left(\frac{2\pi j}d\right)^2+q\right)^m}\,_2F_1\left(m,1-n;2-m-n;\frac{4\pi j}{2\pi j+id\sqrt q}-1\right)}$$
where if the indices’ upper bound is $\infty$, the sums are interchangeable. As $j\to\infty$, the sum converges more gradually.