Nondimensionalizing Fourth Order Differential Equation for an Elastic Beam Under Tension

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I am going through the textbook A First Look At Perturbation Theory 2nd ed. by James G. Simmonds and James E. Mann Jr.

Exercise 1.14 states: "An elastic beam of section modulus $EI$, resting on an elastic foundation of modulus $k$, is under a tension $T$ and a distributed downward force/length $p(s)$, where $s$ is distance along the beam measured from some convenient point. The small vertical deflection $w$ of the beam satisfies the differential equation $$\frac{d^2}{ds^2}\left(EI\frac{d^2w}{ds^2}\right)-T\frac{d^2w}{ds^2}+kw=p$$ For simplicity, assume that $EI$, $T$, and $k$ are constants, expressed in some common set of physical units. Show, by setting $s=Lx$ and appropriately choosing $L$ (which has dimensions of length), that the differential equation can be given the dimensionless form $$\epsilon^2(y'''' +y)-y''=\beta f(x)$$ where $y'=dy/dx$."

I was hoping someone would be able to help me with this problem. I haven't done very much yet and don't know what I should choose $L$ to be.

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Making $w = y_0y(s)$ and $s=L x$ we have

$$ \frac{y_0EI}{L^4}y^{(4)}+y_0 k y - \frac{y_0T}{L^2}y^{(2)}= d $$

now choosing $L$ such that $\frac{EI}{L^4}= k = \sigma$ and $\frac{y_0T}{L^2}=1$ we have:

$$ y_0\sigma\left(y^{(4)}+y\right)-y^{(2)}=d $$

now calling $y_0\sigma = \epsilon^2$ finally

$$ \epsilon^2\left(y^{(4)}+y\right)-y^{(2)}=d $$