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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?

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How is this "partial"? I only see derivatives with respect to $x$. – Robert Israel Jun 10 '12 at 23:03
    
The differential equation (if that's really what you're after) has general solution $c_1 + c_2 (x-1/a) + c_3 (x-1/a)^2 + c_4 (x-1/a)^5$, where $c_1, \ldots, c_4$ are arbitrary constants. – Robert Israel Jun 10 '12 at 23:08
    
Thank you Mr Israel. that was very helpful. – sani Jun 11 '12 at 8:04
    
That is what I was after. Thank you. – sani Jun 11 '12 at 8:11
up vote 3 down vote accepted

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.

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Thank you Valentin. You are a good teacher. – sani Jun 11 '12 at 8:13

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.

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