2nd order differential equation (nonhomogenous)

j3dwards
Messages
32
Reaction score
0

Homework Statement


Find the general solution f(t) of the differential equation: f''(t) − f'(t) − 12f(t) = 36π . How many free parameters enter the solution, and why?

Homework Equations


f = fH + fP where fH is the homogeneous solution and fP is the particular solution.

The Attempt at a Solution


So I've solved this equation to get: f(t) = C1e4t + C2e-3t which I have checked is right.

My question is how may free parameters enter the solution? I'm not sure what this means - is it asking how many constants there are and the answer would therefore be 2? But I don't know why.

Please help! Thank you.
 
Physics news on Phys.org
First, why in the world is a question about "differential equations" posted in the "PreCalculus" section?

Yes, the "free parameters" refer to the number of arbitrary constants. Further the basic theory here is that "the set of all solutions to an nth order linear homogeneous differential equation form a vector space of dimension 2". This is a second order differential equation so the dimension is 2- any solution can be written as a linear combination of two independent solutions giving the two constants.
 
j3dwards said:

Homework Statement


Find the general solution f(t) of the differential equation: f''(t) − f'(t) − 12f(t) = 36π . How many free parameters enter the solution, and why?

Homework Equations


f = fH + fP where fH is the homogeneous solution and fP is the particular solution.

The Attempt at a Solution


So I've solved this equation to get: f(t) = C1e4t + C2e-3t which I have checked is right.

My question is how may free parameters enter the solution? I'm not sure what this means - is it asking how many constants there are and the answer would therefore be 2? But I don't know why.

Please help! Thank you.
It may be a question of terminology, but I have seen the constants be called the free parameters.
So you have two.

You have not solved the differential equation yet, you have found fH.
You still need to find one fP (shouldn't be too difficult)
 
Samy_A said:
It may be a question of terminology, but I have seen the constants be called the free parameters.
So you have two.

You have not solved the differential equation yet, you have found fH.
You still need to find one fP (shouldn't be too difficult)

Oh I forgot to add in the fP = -3π. So why are there 2 free parameters, do you know? Is it because the equation is twice differentiated leaving two constants? Thank you
 
j3dwards said:
Oh I forgot to add in the fP = -3π. So why are there 2 free parameters, do you know? Is it because the equation is twice differentiated leaving two constants? Thank you
Yes, the solution space of your second order linear homogeneous differential equation has dimension 2.
 
j3dwards said:

Homework Statement


Find the general solution f(t) of the differential equation: f''(t) − f'(t) − 12f(t) = 36π . How many free parameters enter the solution, and why?

Homework Equations


f = fH + fP where fH is the homogeneous solution and fP is the particular solution.

The Attempt at a Solution


So I've solved this equation to get: f(t) = C1e4t + C2e-3t which I have checked is right.

My question is how may free parameters enter the solution? I'm not sure what this means - is it asking how many constants there are and the answer would therefore be 2? But I don't know why.

Please help! Thank you.

Because the DE involves second-order derivatives you need two additional conditions in order to fully express the solution (besides the DE itself). For example, you need to know f(0) and f'(0) [initial-value problem] or f(t1) and f(t2) [a two-point boundary value problem], or some initial value such as f(t1) and large-t-behavior (at t →∞), etc. If I tell you the value of f(0), for example, you will still have one undetermined parameter in the solution, and that parameter may be used to match additional problem information.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top