Homogeneous Second Order O.D.E Problem Please help

In summary: A=120 and B=200. just sub those conditions into v(r)=A+B/r separately and solve A and B simultaneously
  • #1
i_hate_math
150
2

Homework Statement


The electric potential energy v(r) of a charged particle located between two uniformly charged concentric spheres with radii r1 and r2 satisfies the second order differential equation

rv′′+2v′=0, r1≤r≤r2
where r is the distance of the charged particle from the common centre of the spheres.

(a) Determine the general solution of the differential equation, by trying a solution of the form v(r)=rm, m∈R.

(b) Using your answer to part (a), find the electric potential energy of a charged particle between two concentric spheres with radii r1 = 2 cm and r2 = 20 cm, kept at potentials v1 = 220 Volts and v2 = 130 Volts respectively.

Homework Equations


rv′′+2v′=0

The Attempt at a Solution


I tried to use v(r)=rm as a solution:
v'(r)=mrm
v''(r)=m2rm
and sub back in:
m2rm+1+2mrm=0
this yields:
m(mr+2)=0, which leads me no where near the solution

Please help
 
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  • #2
You should rethink what v'(r) and v''(r) are equal to.
 
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  • #3
vela said:
You should rethink what v'(r) and v''(r) are equal to.
Darn it I took the exponential. I'll rework my solution
 
  • #4
i found the general solution to be v(r)=A+Be^(-r). then, what's next?
 
  • #5
blerghh said:
i found the general solution to be v(r)=A+Be^(-r). then, what's next?
Apply these conditions:
i_hate_math said:
two concentric spheres with radii r1 = 2 cm and r2 = 20 cm, kept at potentials v1 = 220 Volts and v2 = 130 Volts respectively.
and you get A=200 and B=120
 
  • #6
i_hate_math said:
Apply these conditions:

and you get A=200 and B=120
is there any other way to simplify the exponential before applying the conditions?
 
  • #7
nawhh i just got the answer! thank you!
 
  • #8
i_hate_math said:
Apply these conditions:

and you get A=200 and B=120
wait howd u get the exponential? the solution i got was v=A/r+B
 
  • #9
i_hate_math said:
wait howd u get the exponential? the solution i got was v=A/r+B
yeah, i was wrong at first. no exponential. you are right :)
 
  • #10
isn't the general solution supposed to be v(r)=A+B/r? because both r^0 and r^-1 satisfy the DE
 
  • #11
oops you guys updated... yeah, and after getting v(r)=A+B/r, how do i sub in the conditions... the conditions doesn't make sense to me
 
  • #12
crazy too said:
oops you guys updated... yeah, and after getting v(r)=A+B/r, how do i sub in the conditions... the conditions doesn't make sense to me
you will eventually get A=120 and B=200. just sub those conditions into v(r)=A+B/r separately and solve A and B simultaneously
 
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  • #13
ohhhhh okay, i got the answer! even tho it doesn't make much sense... thank you for your help!
 
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Related to Homogeneous Second Order O.D.E Problem Please help

1. What is a homogeneous second order O.D.E problem?

A homogeneous second order O.D.E problem is a type of ordinary differential equation where the highest derivative of the dependent variable is squared and all other terms in the equation are linear. It is called "homogeneous" because all the terms in the equation have the same degree.

2. How do you solve a homogeneous second order O.D.E problem?

To solve a homogeneous second order O.D.E problem, you can use the method of undetermined coefficients or variation of parameters. Both methods involve finding a particular solution and a complementary solution, and then combining them to get the general solution.

3. What is the difference between a homogeneous and non-homogeneous second order O.D.E problem?

A homogeneous second order O.D.E problem has all terms in the equation with the same degree, while a non-homogeneous problem has at least one term with a different degree. This difference affects the methods used to solve the equation, as a non-homogeneous problem requires the use of the method of undetermined coefficients.

4. Why are homogeneous second order O.D.E problems important in science?

Homogeneous second order O.D.E problems are important in science because they can model many physical phenomena, such as motion, oscillations, and electrical circuits. They also have applications in engineering, physics, and chemistry.

5. Can a homogeneous second order O.D.E problem have a unique solution?

Yes, a homogeneous second order O.D.E problem can have a unique solution. However, it can also have an infinite number of solutions, depending on the initial conditions given in the problem. The general solution of a homogeneous second order O.D.E problem contains two arbitrary constants, which can be determined by the initial conditions to get a unique solution.

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