The differential equation, $$(1-\alpha x)\partial_x^4 y +2\alpha\partial_x^3 y = 0,$$ is despite the usage of partial derivatives an ordinary differential equation since the function $y=y(x)$ to be determined depends solely on the variable $x$. Introducing the function, $$z=\partial_x^3y,$$ the differential equation is recast in the form of a homogeneous first-order ordinary differential equation, $$(1-\alpha x)\partial_x z +2\alpha z=0.$$ This differential equation can be brought - using physical notation - in the form, $$\frac{dz}{z}=-\frac{2\alpha dx}{1-\alpha x}.$$ Integrating, one obtains, $$\partial_x^3y(x)=z(x)=z(x=0)\exp\left(2\log\vert 1-\alpha x\vert\right)=(\partial_x^3y)_0(1-\alpha x)^2=(\partial_x^3y)_0(\alpha x-1)^2,$$ using $\log 1 = 0$ during the integration. Integrating thrice in order to obtain $y(x)$, one has, $$y(x)=(\partial^3_x y)_0\dfrac{1}{3\alpha}\frac{1}{4\alpha}\frac{1}{5\alpha}(\alpha x-1)^5+c_2x^2 + c_1x+c_0=\dfrac{(\partial_x^3y)_0(\alpha x-1)^5}{60\alpha^3}+c_2x^2+c_1x^1+c_0.$$ From the boundary conditions at $x=0$ one has $c_0=0=c_2$ by inspection. The boundary conditions at $x=z$ could also be applied, but I don't understand why there is a differential equation as a boundary condition. Shall one solve this differential equation first and then evaluate at that point? A convention I am not familiar with?
Although the question is old, perhaps the answer is helpful.
Best regards.