- #1
JulieK
- 50
- 0
I have the following differential equation which I obtained from Euler-Lagrange
variational principle
[itex]\frac{\partial}{\partial x}\left(\frac{1}{\sqrt{y}}\frac{dy}{dx}\right)=0[/itex]
I also have two boundary conditions: [itex]y\left(0\right)=y_{1}[/itex] and
[itex]y\left(D\right)=y_{2}[/itex] where [itex]D[/itex], [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are known
numbers.
I assume that I should integrate once to get
[itex]\frac{1}{\sqrt{y}}\frac{dy}{dx}=f\left(y\right)[/itex]
where [itex]f\left(y\right)[/itex] is a function of [itex]y[/itex]. I would like to get
[itex]y[/itex] as a function of [itex]x[/itex] but the problem is that I don't know the
form of [itex]f\left(y\right)[/itex].
My question, how to solve this equation to get [itex]y[/itex] as a function
of [itex]x[/itex]. Is it possible to guess the form of [itex]f\left(y\right)[/itex], e.g.
from the bounday conditions.
variational principle
[itex]\frac{\partial}{\partial x}\left(\frac{1}{\sqrt{y}}\frac{dy}{dx}\right)=0[/itex]
I also have two boundary conditions: [itex]y\left(0\right)=y_{1}[/itex] and
[itex]y\left(D\right)=y_{2}[/itex] where [itex]D[/itex], [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are known
numbers.
I assume that I should integrate once to get
[itex]\frac{1}{\sqrt{y}}\frac{dy}{dx}=f\left(y\right)[/itex]
where [itex]f\left(y\right)[/itex] is a function of [itex]y[/itex]. I would like to get
[itex]y[/itex] as a function of [itex]x[/itex] but the problem is that I don't know the
form of [itex]f\left(y\right)[/itex].
My question, how to solve this equation to get [itex]y[/itex] as a function
of [itex]x[/itex]. Is it possible to guess the form of [itex]f\left(y\right)[/itex], e.g.
from the bounday conditions.