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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

1 Answer 1

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

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