# [SOLVED]Euler Lagrange equations

#### dwsmith

##### Well-known member
Given this $$F = p(x)y^{'2}-q(x)y^2+2f(x)y$$. What would be the integral of $$f(x)$$ and $$q(x)$$?
\begin{align*}
f(x) - q(x)y - \frac{d}{dx}\left[p(x)y'\right] &= 0\\
\frac{d}{dx}\left[p(x)y'\right] &= f(x) - q(x)y\\
y'p(x) &= \int f(x)dx - y\int q(x)dx
\end{align*}

Last edited:

#### Ackbach

##### Indicium Physicus
Staff member
Assuming you meant to apply the EL equations to $F$, you've started out correctly. I'm not sure you can pull $y$ out of the last integral, though. If $y=y(x)$, then I think you have to have
$$y'p(x) = \int f(x) \,dx - \int y \, q(x) \,dx.$$
The question, "What would be the integral of $f(x)$ and $q(x)$?" is a bit unclear. You have sort-of found it, I suppose, although the $y$ under the $q$ integral is a bit troubling. Is that the exact wording of the original question?

#### dwsmith

##### Well-known member
Assuming you meant to apply the EL equations to $F$, you've started out correctly. I'm not sure you can pull $y$ out of the last integral, though. If $y=y(x)$, then I think you have to have
$$y'p(x) = \int f(x) \,dx - \int y \, q(x) \,dx.$$
The question, "What would be the integral of $f(x)$ and $q(x)$?" is a bit unclear. You have sort-of found it, I suppose, although the $y$ under the $q$ integral is a bit troubling. Is that the exact wording of the original question?

How can I continue to solve the problem is the real question since I have this integrals. I suppose, for f, I could say \int f = F but then I still have the issue of the other integral of yq(x).

#### Ackbach

##### Indicium Physicus
Staff member
Yes, you have an integro-differential equation for $y$. Not so nice. I might actually peal back your integral and try to solve
$$f(x) - q(x)y - p'(x)y' -p(x)y''= 0,$$
or
$$p(x)y''+p'(x)y'+q(x)y=f(x).$$
It's at least linear in $y$. The homogeneous equation, if I'm not mistaken, is Sturm-Liouville. That might help you some.

#### dwsmith

##### Well-known member
This may be helpful then. I know F is integrable on from a to b.
$$\int_a^b Fdx$$
So could one say then $y(a) = y_1$ and $y(b) = y_2$?

#### dwsmith

##### Well-known member
If we solve that DE, here is what I came up with using variation of parameters
\begin{align*}
y'' + \frac{p'}{p}y' + \frac{q}{p}y &= \frac{f}{p}\\
y'' + Ay' + By &= C
\end{align*}
Then we have $$m^2 + Am + B = 0$$ where $$m = \frac{-A\pm\sqrt{A^2-4B}}{2}$$.
Then we have 3 cases:
If $$A^2-4B = 0$$, then
$$y_c = c_1\exp\left[-\frac{A}{2}x\right] + c_2x\exp\left[-\frac{A}{2}x\right]$$
Then the Wronskian is
$$W = \exp(-Ax),$$
$$u_1 = \frac{2C}{A^2}\exp\left[\frac{A}{2}x\right](2-Ax)$$, and $$u_2 = \frac{2C}{A}\exp\left[\frac{A}{2}x\right]$$.
$$y = c_1\exp\left[-\frac{A}{2}x\right] + c_2x\exp\left[-\frac{A}{2}x\right] + \frac{2C}{A^2}\exp\left[\frac{A}{2}x\right](2-Ax) + \frac{2C}{A}\exp\left[\frac{A}{2}x\right]$$

If $$A^2-4B < 0$$, then
$$y_c = \exp\left[-\frac{A}{2}x\right]\left(c_1\cos\left(-\frac{\sqrt{A^2-4B}}{2}x\right) + c_2\sin\left(-\frac{\sqrt{A^2-4B}}{2}x\right)\right)$$
Repeat first case here:

If $$A^2-4B > 0$$, then
$$y_c = c_1\exp\left[\frac{-A+\sqrt{A^2-4B}}{2}\right] + c_2\exp\left[\frac{-A-\sqrt{A^2-4B}}{2}\right]$$
Repeat first case here:

[HR][/HR]
Maybe I misunderstood the original question though since this is rather intensive.

It original said:
Obtain the E-L eqs associated with extremizing $$\int_a^bFdx$$.

#### Ackbach

##### Indicium Physicus
Staff member
I'm afraid your solution to the DE is not valid. The guess-and-check methods associated with the exponential only work with constant coefficients.

Given your problem statement, I think you could stop here:
$$p(x)y''+p'(x)y'+q(x)y=f(x).$$
This DE is the result of applying the EL equation to $F$.