- #1

- 2

- 0

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

Can someone help with this?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter shuttleman11
- Start date

- #1

- 2

- 0

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

Can someone help with this?

- #2

rock.freak667

Homework Helper

- 6,223

- 31

d

- #3

- 2

- 0

- #4

rock.freak667

Homework Helper

- 6,223

- 31

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

Science Advisor

Homework Helper

- 41,847

- 966

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.

Share: