is it possible to solve an equation with the given form analytically?
$$ (1-ax^2)*e^{bx^2}=c $$ $$ e^{bx^2}-ax^2e^{bx^2}=c $$
I've already tried it using a logarithmic function but I cannot manage it to get $x$ on one site.
I need to get this equation to the generalized transcendental algebraic equation: $$ e^{-cx} = a (x-b) $$
in order to solve it with $x=b+\frac{1}{c}W(c \frac{e^{-cb}}{a})$ where $W$ is theLambert function.
Am I right?
Thanks a lot!
There was a question similar to this asked yesterday. See my answer there. The solution, to the extent there is one, involves the Lambert W function.
Added later: Here are the steps I followed to relate the OP's equation to the W function. First, multiply both sides by $-b/a$ to get
$$-{bc\over a}=\left(bx^2-{b\over a} \right)e^{bx^2}$$
Next multiply both sides by $e^{-{b\over a}}$ to get
$$-{bc\over a}e^{-{b\over a}}=\left(bx^2-{b\over a} \right)e^{bx^2-{b\over a}}$$
This is now of the form $z=We^W$, where $z=(-bc/a)e^{-b/a}$ and $W=bx^2-(b/a)$. Hence
$$x=\sqrt{W\left(-{bc\over a}e^{-b/a}\right)+{b\over a}\over b}$$
If there were some other, elementary expression for $x$ as a function of $a$, $b$, and $c$, then you would be able to unwind things and get an elementary expression for the W function.