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Find the Euler Equation

Problem Statement
Write and solve the Euler equation to make the integral equation stationary.
Relevant Equations
∫(y'^2+y^2)dx
∫(y'^2+y^2)dx

Why I obtain two different equations?

1. y''=y

2. y'-xy+C=0
 
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Problem Statement: Write and solve the Euler equation to make the integral equation stationary.
Relevant Equations: ∫(y'^2+y^2)dx

∫(y'^2+y^2)dx
This is not an equation -- there's no = sign.

You need to provide more context. As an ordinary differential equation, an Euler equation looks like this:
##ax^2y'' + bxy' + c = 0##
See http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
There are also Euler equations that describe fluid flow.
So what kind of Euler equation are you trying to solve?
Fernando Rios said:
Why I obtain two different equations?

1. y''=y

2. y'-xy+C=0
No idea, since you didn't show what you did.
 

Orodruin

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You need to provide more context.
The context is ”to make the integral stationary”. He is looking for the Euler-Lagrange equation, by some authors just called Euler equation.

Problem Statement: Write and solve the Euler equation to make the integral equation stationary.
Relevant Equations: ∫(y'^2+y^2)dx

Why I obtain two different equations?

1. y''=y

2. y'-xy+C=0
You do not. If you do it correctly, the first integral will be the integrated version of the EL equation. That is where the integration constant comes from.

You should show us how you arrived at those two equations if you want further help than that.
 
The context is ”to make the integral stationary”. He is looking for the Euler-Lagrange equation, by some authors just called Euler equation.


You do not. If you do it correctly, the first integral will be the integrated version of the EL equation. That is where the integration constant comes from.

You should show us how you arrived at those two equations if you want further help than that.
I start with the following:
d/dx(dF/dy')-dF/dy=0

d/dx(d/dy'(y'^2+y^2))-d/dy(y'^2+y^2)=0

d/dx(2y')-2y=0

2d/dx(y')-2y=0

d/dx(y')-y=0

First path and the one found in the solution:
y''=y

Second path:
∫d(y')=∫ydx

y'=xy+C

What is wrong with the second path?
 
The first solution is perfectly fine, to understand why the second is wrong you have to keep in mind that ##y## is a function of ##x## so $$\int y(x) \text{d}x\neq yx+C$$. To find ##y(x)## you have to solve the equation you get: ##\frac{d^2y(x)}{dx^2}=y(x)## that is an easy differential equation.
 
The first solution is perfectly fine, to understand why the second is wrong you have to keep in mind that ##y## is a function of ##x## so $$\int y(x) \text{d}x\neq yx+C$$. To find ##y(x)## you have to solve the equation you get: ##\frac{d^2y(x)}{dx^2}=y(x)## that is an easy differential equation.
You are totally right. I don't know why I didn't notice that before. Thanks a lot.
 

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