Question 798503
That's a lot to write, so I will define
{{{Y(x,t)=1/(x+Lt+ a)=(x+Lt+ a)^(-1)}}}
 
Now {{{w(x,t)= L + 2Y}}} is much easier to write.
 
{{{dw/dt= (dw/dY)(dY/dt)= 2(-1(x+Lt+ a)^(-2)*L)=-2LY^2}}}
 
{{{dw/dx=(dw/dY)(dY/dx)=2(-1(x+Lt+ a)^(-2))=-2(x+Lt+ a)^(-2)=-2Y^2}}}
 
{{{d^2w/dx^2=d(dw/dx)/dx=d(-2Y^2)/dx=(d(-2Y^2)/dY)(dY/dx)=(-2(2Y))(-1(x+Lt+ a)^(-2))=(-4Y)(-Y^2)=4Y^3}}}
 
That was the calculus part.
The rest is just algebra:
{{{d^2w/dx^2+ w(dw/dx)=4Y^3+w(-2Y^2)=4Y^3+(L+2Y)(-2Y^2)=4Y^3-2LY^2-4Y^3=-2LY^2=dw/dt}}}