finding solution to partial differential equation what is the best way to solve a partial differential equation:
$$
(1-ax)(∂^4 y)/(∂x^4)+2a (∂^3 y)/(∂x^3)=0
$$
like in ordinary differential equations I tried the power series method (I'm not very good with differential equations). I got something like:
$$y= C_1+C_2+C_3 (1+(1/3) ax)+C_4 (1-ax)$$
which is difficult subjecting to the boundary conditions:
$$
y=0,y''=0,x=0 \\
y=M,y''=(-1-y')/k(1-ax),x=z
$$
Can anyone help?
 A: Since the statement of the problem gives no reason to consider the equation a PDE, a standard method of lowering the order applies:
$$(1-ax)\frac{d^4y}{dx^4} +2a\frac{d^3y}{dx^3}=0$$
$$z=\frac{d^3y}{dx^3}$$
$$(1-ax)z'+2az=0$$
$$\frac{z'}{z}=-\frac{2a}{1-ax}$$
$$\ln |z|=\ln[(1-ax)^2]+C_1$$
$$z=C_1(1-ax)^2$$
Now integrate three times and apply boundary conditions as appropriate.
A: 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.
