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

Can someone help with this?

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- Thread starter shuttleman11
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- #1

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

Can someone help with this?

- #2

Homework Helper

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d

- #3

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- #4

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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

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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?

And what is [itex]\int d^2y/dx^2 dx[/itex]?

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.

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