The law states that the wavelength where the energy density  is maximum. λmax and product of the T is a constant

$$λ_{max} T = 0.289 cm^1K$$

To find the maximum of u (v, T)

$$\frac{du}{dv}=\frac{8\pi h}{c^3}\left[\frac{v^3}{e^{\displaystyle\frac{h\nu}{kT}}-1}\right]$$

For u to be maximum $\frac{du}{dv}=0$

The above equation becomes

So,

$$3\nu^2\left(e^\frac{h\nu}{kT}-1\right)=\nu^3\frac h{kT}e^\frac{h\nu}{kT}$$

Let $$ \frac{h\nu}{kT}=\frac{hc}{\lambda kT}=x$$

The above equation become.

$$e^{-x}+\frac x5=1$$

A numerical solution gives $x=4.965 $

$λ_{max} T=\frac{hc}{ kT4.965}=0.288 cm^-1K$

This is the Wein’s displacement formula.