Need help on ordinary differentail equation problem

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In summary, the students were trying to find potential in a cube using Laplace's equation and found that f(x)+g(y)+h(z) must be constant for this to be true. They found the c1, c2, and c3 constants and solved an ODE to get X(x)=Asin(ky)+bcos(ky).
  • #1
leonne
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Homework Statement


Well the problem is a electromagnetism physics problem to find potential in a cube.


Homework Equations





The Attempt at a Solution


using Laplace question and combining X(x)Y(y)Z(y) to it
we get
1/X(d2X/dx2 + 1/Y(d2Y/dy2 + 1/Y(d2Y/dy2=0

Than we see its in the forum of f(x)+ g(y)+ h(z)=0 that they must be constant for this to be true so
1/X(d2X/dx2=c1 1/Y(d2Y/dy2=c2
1/Y(d2Y/dy2=c3

so c1+c2+c3=0 They go about saying like how c3 is pos while c1 ,c2 is negative
( btw do u know why its that? some other problems has c1 pos while c2 c3 neg does it matter?)

Than they have c1=-k2 c2=-l2 than c3=(k+l)2
btw where did the k and l come from is it just some constant they picked?

now the plug in for the c we found and get like (d2X/dx2=-k2 X...
Than they solve this ODE which is the part i am lost. They just go from this to having
X(x)=Asin(ky)+bcos(ky) why is that? also in another problem when X was pos it was Aekx+Be-kx why is it like this?

For this prblem the pos z was in the same format as that. Should i just accept that when its pos to use e^ and when neg to use the cos sine?
 
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  • #2
leonne said:

Homework Statement


Well the problem is a electromagnetism physics problem to find potential in a cube.


Homework Equations





The Attempt at a Solution


using Laplace question and combining X(x)Y(y)Z(y) to it
we get
1/X(d2X/dx2 + 1/Y(d2Y/dy2 + 1/Y(d2Y/dy2=0

Than we see its in the forum of f(x)+ g(y)+ h(z)=0 that they must be constant for this to be true so
1/X(d2X/dx2=c1 1/Y(d2Y/dy2=c2
1/Y(d2Y/dy2=c3

so c1+c2+c3=0 They go about saying like how c3 is pos while c1 ,c2 is negative
( btw do u know why its that? some other problems has c1 pos while c2 c3 neg does it matter?)
That does NOT follow from what you give here. It might follow from the boundary conditions you are given for the problem.

Than they have c1=-k2 c2=-l2 than c3=(k+l)2
btw where did the k and l come from is it just some constant they picked?
Basically, yes. Since a square is always positive, writing c1 and c2 as "[itex]-k^2[/itex]" and "[itex]-l^2[/itex]" emphasizes that they are negative. Also, it simplifies the solution to the "characteristic equation".

now the plug in for the c we found and get like (d2X/dx2=-k2 X...
Than they solve this ODE which is the part i am lost. They just go from this to having
X(x)=Asin(ky)+bcos(ky) why is that? also in another problem when X was pos it was Aekx+Be-kx why is it like this?
The "characteristic equation" for the differential equation ay"+ by+ c= 0 is [itex]ar^2+ br+ c= 0[/itex]. If that has two distinct real solutions, say [itex]r_1[/itex] and [itex]r_2[/itex], then the general solution is [itex]y(t)= Ce^{r_1t}+ De^{r_2t}[/itex]. If it has two complex roots, a+ bi and a- bi, then the general solution is [itex]y(t)= e^{at}(Ccos(bt)+ Dsin(bt))[/itex]. The two different kinds of solutions are related through [itex]e^{iat}= cos(at)+ i sin(at)[/itex]

For this prblem the pos z was in the same format as that. Should i just accept that when its pos to use e^ and when neg to use the cos sine?
Unfortunately, one of the basic rules of physics/mathematics is "you always learn the mathematics you need for physics the next year!". It looks to me like this course is using differential equations intensively and differential equations should have been a prerequisite for it.

All I can say is that if you are in a course in which you are apparently expected to solve partial differential equations and you have not yet taken Ordinary Differential Equations, you are going to have to accept a lot of things without really understanding them. You might try looking at http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/lecture-notes/
 
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  • #3
l thxs for the info yeah you only needed calc 3 for this course. I already took ode but like forgot everything what i learned lol. This was like the only section that uses ode, but would be nice if they gave a little detail in the book lol well thanks
 

FAQ: Need help on ordinary differentail equation problem

What is an ordinary differential equation (ODE)?

An ordinary differential equation is a mathematical equation that describes how a variable changes over time based on its current value and its rate of change. It involves derivatives, which represent the rate of change of a variable, and the variable itself.

How do I solve an ODE?

The solution to an ODE involves finding a function that satisfies the equation. This can be done analytically or numerically. Analytical solutions use mathematical techniques such as separation of variables or substitution to find an exact solution. Numerical solutions use algorithms to approximate the solution.

What is the order of an ODE?

The order of an ODE refers to the highest derivative present in the equation. For example, a first-order ODE would have only one derivative, while a second-order ODE would have a second derivative.

What are some real-life applications of ODEs?

ODEs are used to model a wide range of phenomena in science and engineering, such as population growth, chemical reactions, and motion of objects under the influence of forces. They are also used in economics, finance, and biology.

How can I check if my solution to an ODE is correct?

To check the correctness of an analytical solution, you can substitute it back into the original equation and see if it satisfies the equation. For numerical solutions, you can compare it to other numerical methods or use known solutions to test its accuracy.

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