# Solve differential equation with boundary conditions using substitution

cheesecake91
μ$^{2}$$\frac{d^{2}u}{dx^{2}}$+ae$^{u}$=0

Boundary conditions: u(-L)=u(L)=u$_{0}$

Solve by multiplying by $\frac{du}{dx}$ and integrating in x

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

## Answers and Replies

danielu13
μ$^{2}$$\frac{d^{2}u}{dx^{2}}$+ae$^{u}$=0

Boundary conditions: u(-L)=u(L)=u$_{0}$

Solve by multiplying by $\frac{du}{dx}$ 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
$\mu$$^2$ $'' = -ae^\mu$

Then you solve for a general solution which is
$\mu(x) = 1 + x$

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:
$\frac{\mu^4}{12} = -ae^\mu + \mu*c_1 + c_2$

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

Last edited:
JJacquelin
Hi !
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).