Solve the equation $-Axe^{-2x^{2}}+Bxe^{-x^{2}}+Cx+D=0$ where $A,B,C\geq 0$ and $D\in\mathbb{R}$.

Question

Solve the following equation:

$$-Axe^{-2x^{2}}+Bxe^{-x^{2}}+Cx+D=0$$ where $A,B,C\geq 0$ and $D\in\mathbb{R}$.

Remark: The difficulty occurs when we consider $D\neq 0$. I do not know how to determine a solution of this equation.

Answer 1

A "closed-form" solution is extremely unlikely, even if $A=0$ or $B=0$ or $C=0$. Numerical methods can be used. Or you could try series solutions: for small $D$,

$$ \left( A-B-C \right) ^{-1}D+{\frac {2\,A-B}{ \left( A-B-C \right) ^{ 4}}}{D}^{3}+{\frac {20\,{A}^{2}-19\,AB+4\,AC+5\,{B}^{2}-BC}{2\, \left( A-B-C \right) ^{7}}}{D}^{5}+{\frac {392\,{A}^{3}-545\,{A}^{2}B +176\,{A}^{2}C+274\,A{B}^{2}-126\,ABC+8\,A{C}^{2}-49\,{B}^{3}+22\,{B}^ {2}C-B{C}^{2}}{6\, \left( A-B-C \right) ^{10}}}{D}^{7}+{\frac {11664\, {A}^{4}-21311\,{A}^{3}B+8368\,{A}^{3}C+15619\,{A}^{2}{B}^{2}-9779\,{A} ^{2}BC+1072\,{A}^{2}{C}^{2}-5381\,A{B}^{3}+3914\,A{B}^{2}C-589\,AB{C}^ {2}+16\,A{C}^{3}+729\,{B}^{4}-523\,{B}^{3}C+67\,{B}^{2}{C}^{2}-B{C}^{3 }}{24\, \left( A-B-C \right) ^{13}}}{D}^{9}+O \left( {D}^{10}\right) $$

Answer 2

An exact solution doesn't exist, unless you provide more information about the constants, which may turn the equation into something knows.

For a general case, we have different cases with respect upon what $x$ is.

The first simple case is when $x$ is a quite big number, which doesn't mean thousands or hundreds but also few tens, or even $x > 3$ may hold. The reason is that in this case we can expand the exponential in Taylor series remembering

$$e^z \approx 1 + z + \frac{1}{2}z^2 + \mathcal{O}(z^3)$$

We will take, however, only the first two terms for each exponential:

$$e^{-2x^2} \approx 1 - 2x^2$$

$$e^{-x^2} \approx 1 - x^2$$

Substituting into the equation, and collecting the powers of $x$ we end up with

$$^3(2A - B) + x(B - A + C) + D = 0$$

General case

Assuming $2A - b \neq 0$ we can call it $p$ and divide by p. We will call then $q = B - A + C$ and $t = q/p$ whilst $u = d/p$, getting:

$$x^3 + tx + u = 0$$

Which is a cubic equation special edition because the $x^2$ term is missing.

Such an equation has solutions by Cardano's simplified formula:

$$x\to \frac{\sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}{\sqrt[3]{2} 3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}} t}{\sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}$$

$$x\to \frac{\left(1+i \sqrt{3}\right) t}{2^{2/3} \sqrt[3]{3} \sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}{2 \sqrt[3]{2} 3^{2/3}}$$

$$x\to \frac{\left(1-i \sqrt{3}\right) t}{2^{2/3} \sqrt[3]{3} \sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{3} \sqrt{4 t^3+27 u^2}-9 u}}{2 \sqrt[3]{2} 3^{2/3}}$$

Which you can obtain indeed by using Cardano's resolvent.

Second case $2A - B = 0$

Clearly here we cannot call it $p$ and divide by $p$. In this case we recognize simply to be $ A = B/2$ hence the equation becomes linear:

$$x\left(\frac{B}{2} + C\right) + D = 0$$

With a simple solution

$$x = -\frac{D}{\frac{B}{2} + C}$$

Notice that both the cases above get really simplified in the case $D = 0$.

Other possibilities

Other methods for example include taking more terms in the expansion, in which we shall then include respectively for the two exponentials:

$$ + 2x^4$$

$$ + \frac{1}{2}x^4$$

This will turn the equation into a quartic, which is actually another solvable equation with Ferrari's methods I presented a little time ago in this answer: {https://math.stackexchange.com/q/2649336}

Taking more orders will lead you to better approximations but it will become unhandy / almost improbable to solve it.