- #1

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Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

I know you have to use substitution, but I keep going in circles.

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- Thread starter cheesecake91
- Start date

- #1

- 1

- 0

Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

I know you have to use substitution, but I keep going in circles.

- #2

- 67

- 0

Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

I know you have to use substitution, but I keep going in circles.

The way that I would do it would be to rewrite the equation as

[itex]\mu[/itex][itex]^2[/itex] [itex]'' = -ae^\mu[/itex]

Then you solve for a general solution which is

[itex]\mu(x) = 1 + x[/itex]

Then you go on to get your general solution from the boundaries and such.

Another method, which would probably be easier and is what I think you were trying to do is to just integrate twice and get:

[itex]\frac{\mu^4}{12} = -ae^\mu + \mu*c_1 + c_2[/itex]

That may be slightly off, but try integrating it yourself, and then use your boundary conditions to solve for the constants.

Last edited:

- #3

- 798

- 34

I think that the two methods proposed by danielu13 are both incorrect.

Better follow the advice given in the wording : multiply by u'

m² u'' +a exp(u) = 0

m² 2 u''u' +2a exp(u) u' = 0 and integrate :

m² u'² +2a exp(u) = c

m u' = (+ or -)sqrt(c-2a exp(u))

Case +sqrt :

m (du/dx) = sqrt(c-2a exp(u))

dx = m du / sqrt(c-2a exp(u))

Then integrate, wich gives x(u)

and invert it to obtain u(x).

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