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

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?

LCKurtz
Homework Helper
Gold Member
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?

HallsofIvy
Homework Helper
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.

LCKurtz
Homework Helper
Gold Member
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).

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!

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!