2nd order differential equation (nonhomogenous)

In summary, a 2nd order differential equation is a mathematical equation that describes the change of a variable over time. It is considered nonhomogeneous when it includes terms or functions that are not equal to zero, and can be solved using methods such as the method of undetermined coefficients, variation of parameters, or Laplace transform. 2nd order differential equations are crucial in various fields of science, as they are used to model and predict the behavior of physical systems. They can have more than one solution, depending on the initial conditions and the values of the arbitrary constants in the general solution.
  • #1
j3dwards
32
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.
 
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  • #2
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.
 
  • #3
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)
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 

FAQ: 2nd order differential equation (nonhomogenous)

1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that describes how a variable changes over time, based on its current value and the rate at which it is changing. It involves the second derivative (rate of change of the rate of change) of the variable.

2. What makes a differential equation nonhomogeneous?

A differential equation is considered nonhomogeneous when it includes terms or functions that are not equal to zero. These terms can be constants, variables, or functions of the variable being solved for.

3. How do you solve a nonhomogeneous 2nd order differential equation?

To solve a nonhomogeneous 2nd order differential equation, you can use various methods such as the method of undetermined coefficients, variation of parameters, or Laplace transform. These methods involve finding a particular solution that satisfies the nonhomogeneous term and adding it to the general solution of the corresponding homogeneous equation.

4. What is the importance of 2nd order differential equations in science?

2nd order differential equations are essential in various fields of science, such as physics, engineering, and economics. They are used to model and predict the behavior of physical systems and phenomena, such as motion, oscillations, and population growth.

5. Can a 2nd order differential equation have more than one solution?

Yes, a 2nd order differential equation can have multiple solutions. This is because the general solution of a 2nd order differential equation contains two arbitrary constants, which can take on different values for different initial conditions. However, in some cases, there may be only one unique solution to a particular 2nd order differential equation.

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