Integration by parts of 4th order DE

  • #1
having difficulty integrating the following equation by parts to determine if its symmetric:

d4 u / dx4 + K d2 u / dx2 + 6 = 0 0< x < 1

Can someone help with this?
 

Answers and Replies

  • #2
rock.freak667
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If K is a constant then you have a 4th order DE with constant coefficients. So that all of your answers will be in the form y=erx.

d4u/dx4 + K d2u/dx2 + 6=0
 
  • #3
But I'm lost with assigning u, v, du, dv to this equation? Also can I split it into three idfferent integrals added together or must I assign the values and differentiate the hole equation by parts? Not for sure if I'm being clear?
 
  • #4
rock.freak667
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[tex]\frac{d^4u}{dx^4} + K \frac{d^2u}{dx^2} + 6=0[/tex]

Now you can just integrate everything with respect to x to get

[tex]\int \frac{d^4u}{dx^4}dx + \int K \frac{d^2u}{dx^2}dx + \int 6dx= \int 0dx[/tex]



Now [itex]\int \frac{dy}{dx}dx=y+c[/itex] where c is a constant. Now just use this idea to work out your problem.
 
  • #5
HallsofIvy
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But I'm lost with assigning u, v, du, dv to this equation? Also can I split it into three idfferent integrals added together or must I assign the values and differentiate the hole equation by parts? Not for sure if I'm being clear?
This is NOT a first order equation: you can't just integrate both sides. rockfreak667 told you to do it as an equation with constant coefficients. What is its characteristic equation?

[tex]\frac{d^4u}{dx^4} + K \frac{d^2u}{dx^2} + 6=0[/tex]

Now you can just integrate everything with respect to x to get

[tex]\int \frac{d^4u}{dx^4}dx + \int K \frac{d^2u}{dx^2}dx + \int 6dx= \int 0dx[/tex]


Now [itex]\int \frac{dy}{dx}dx=y+c[/itex] where c is a constant. Now just use this idea to work out your problem.
And what is [itex]\int d^2y/dx^2 dx[/itex]?
 

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