Can you help with this 2nd-order, linear, homog, w/const.coeff proof?

In summary, if the roots of the characteristic equation are real, then a solution of ay" + by' + cy = 0 is either everywhere zero or can take on the value zero at most once, because the existence and uniqueness theorem for linear equations with constant coefficients shows that the solution is determined by its value at two points and setting the solution equal to zero at two points yields the trivial solution which can only occur when either the coefficients or t are equal to zero.
  • #1
diligence
144
0
Question:

If the roots of the characteristic equation are real, show that a solution of ay" + by' + cy = 0 is either everywhere zero or else can take on the value zero at most once.


Okay, if the roots of the CE are real, then the solution takes on one of two forms:
(i) y(t) = c_1 e^(r_1 t) + c_2 e^(r_2 t)
or
(ii) y(t) = c_1 e^(r_1 t) + c_2 t e^(r_2 t)


So now either both c_1 and c_2 are both zero, making the solution everywhere zero, or else if c_1 and c_2 are not both zero, the solution can only have the zero value at most once. I think this is because the exponential function is either always increasing or always decreasing, or because if you combine two exponential functions that it only changes from increasing to decreasing at most once (is that even correct?)...but i have no idea how to PROVE this!

maybe take critical points of the solution and show that there is at most one critical point?

can anyone help with this?
 
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  • #2
diligence said:
Question:

If the roots of the characteristic equation are real, show that a solution of ay" + by' + cy = 0 is either everywhere zero or else can take on the value zero at most once.Okay, if the roots of the CE are real, then the solution takes on one of two forms:
(i) y(t) = c_1 e^(r_1 t) + c_2 e^(r_2 t)
or
(ii) y(t) = c_1 e^(r_1 t) + c_2 t e^(r_2 t)

In case ii the root is repeated: y(t) = (c1 + c2t)ert

So now either both c_1 and c_2 are both zero, making the solution everywhere zero, or else if c_1 and c_2 are not both zero, the solution can only have the zero value at most once. I think this is because the exponential function is either always increasing or always decreasing, or because if you combine two exponential functions that it only changes from increasing to decreasing at most once (is that even correct?)...but i have no idea how to PROVE this!

maybe take critical points of the solution and show that there is at most one critical point?

can anyone help with this?

In each case, why not try setting them equal to 0 and see how many t values work?
 
  • #3
But a simpler proof is this: any linear equations with constant coefficients (the coefficient of the highest derivative being non-zero, of course) satisfies the existence and uniqueness theorem. In particular, the solution to a second order linear equation with constant coefficients is determined by its value at two points. Since y= 0 for all x satisfies the differential equation, it is the only solution that is 0 at those two points.
 
  • #4
HallsofIvy said:
But a simpler proof is this: any linear equations with constant coefficients (the coefficient of the highest derivative being non-zero, of course) satisfies the existence and uniqueness theorem. In particular, the solution to a second order linear equation with constant coefficients is determined by its value at two points. Since y= 0 for all x satisfies the differential equation, it is the only solution that is 0 at those two points.

No. You are thinking of existence and uniqueness theorem for the initial value problem, which doesn't apply to two point boundary value problems. Look at:

y'' + y = 0, y(0) = y(pi) = 0

which is satisfied by y = sin(x).
 
  • #5
LCKurtz said:
In case ii the root is repeated: y(t) = (c1 + c2t)ert



In each case, why not try setting them equal to 0 and see how many t values work?


Thanks. I was overthinking it. I had set them to zero like you suggested but when i couldn't take the logarithm i gave up, thinking i need to find a new method...but that's the proof right there because it can't equal zero unless the coefficients or t equal zero, which can only happen at most once. So it's either everywhere zero (coeff.=0) or at most equal to zero once (t=0).

Thanks!
 
  • #6
i need to learn to avoid what hindered me in this problem. Specifically, the fact that I immediately discarded a null result, thinking that maybe i did something wrong, when in fact, that is the result!
 

1. What is a second-order linear homogeneous equation with constant coefficients?

A second-order linear homogeneous equation with constant coefficients is a type of differential equation that can be written in the form:
y'' + ay' + by = 0
where y is the unknown function, a and b are constants, and y' and y'' represent the first and second derivatives of y, respectively. The equation is considered homogeneous because all terms involve the unknown function y and its derivatives.

2. What is the purpose of a proof in mathematics?

A proof in mathematics is a rigorous and logical argument that is used to establish the truth of a statement or theorem. It is an essential part of the scientific method, allowing mathematicians to verify the validity of their theories and conclusions.

3. How do you approach a proof for a second-order linear homogeneous equation with constant coefficients?

To prove a second-order linear homogeneous equation with constant coefficients, you can use the method of undetermined coefficients or the method of variation of parameters. Both methods involve substituting an assumed solution into the equation and solving for the unknown coefficients.

4. Can you provide an example of a second-order linear homogeneous equation with constant coefficients?

One example of a second-order linear homogeneous equation with constant coefficients is:
y'' + 4y' + 4y = 0
This equation can be solved using the method of undetermined coefficients, and the general solution is y = c1e^(-2x) + c2xe^(-2x), where c1 and c2 are constants.

5. What are the key properties of a second-order linear homogeneous equation with constant coefficients?

The key properties of a second-order linear homogeneous equation with constant coefficients include linearity, homogeneity, and constancy of coefficients. Linearity means that the equation can be written as a linear combination of the unknown function and its derivatives. Homogeneity means that all terms in the equation involve only the unknown function and its derivatives. Constancy of coefficients means that the coefficients a and b are constants and do not depend on the independent variable.

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