# Convert $(1+2x)^2\frac{d^2y}{dx^2}-6(1+2x)\frac{dy}{dx}+16y=8(1+2x)^2$ to a form with constant coefficient

Convert the differential equation $$(1+2x)^2\frac{d^2y}{dx^2}-6(1+2x)\frac{dy}{dx}+16y=8(1+2x)^2$$ to a form with constant coefficient.

How do we arrive at the following?

Let $1+2x=e^z$

$\therefore z=\log (2x+1)$

So that $(2x+1)\frac{dy}{dx}=2\frac{dy}{dz}=2D'y$

and $(2x+1)^2\frac{d^2y}{dx^2}=2^2.D'(D'-1)y$

This is Euler equation $$(1+2x)^2y''-6(1+2x)y'+16y=8(1+2x)^2$$ with substitution $1+2x=e^z$ or $z=\ln(1+2x)$ where $1+2x>0$ we have $$2\dfrac{dy}{dz}=2\dfrac{dy}{dx}\dfrac{dx}{dz}=2y'\dfrac{e^z}{2}=(1+2x)y'$$ and $$2.2\dfrac{d^2y}{dz^2}=2\dfrac{d}{dz}(2\dfrac{dy}{dz})=2\dfrac{d}{dx}((1+2x)y')\dfrac{dx}{dz}=2\left(2y'+(1+2x)y''\right)\dfrac{e^z}{2}=2(1+2x)y'+(1+2x)^2y''$$ or $$(1+2x)y'=2\dfrac{dy}{dz}~~~~,~~~~(1+2x)^2y''=4\dfrac{d^2y}{dz^2}-4\dfrac{dy}{dz}$$ then we have a new equation respect to $z$ variable: $$y''-4y'+4y=2e^{2z}$$
• The LHS is okay but my RHS is $2e^{2z}$. Are you sure with your RHS? – Fakemistake Aug 13 '18 at 14:29
• You are right. RHS is $2e^{2z}$, thanks. – Nosrati Aug 13 '18 at 16:32