That's what it says... Hey, I don't make the rules.

...

∂w
∂t
+
∂3w
∂x3 + 6¾w2 ∂w
∂x
= 0.
Modified Korteweg–de Vries equation.
1±. One solution on
solution for ¾ = 1:
w(x, t) = a + k2
p4a2 + k2 cosh z + 2a
, z = kx - (6a2k + k3)t + b,
where a, b, and k are arbitrary constants.
2±.
solution for ¾ = 1:
w(x, t) = 2 a1eµ1 + a2eµ2 + Aa2e2µ1+µ2 + Aa1eµ1+2µ2
1 + e2µ1 + e2µ2 + 2(1 - A)eµ1+µ2 + Ae2(µ1+µ2)...

f(xy) = f(x)f(y)

x″ = Fx, y″ = Fy, z″ = Fz, where F = F(r), r = (x2 + y2 + z2)1/2.