Consider the equation: ay'' +by'+cy=0 Consider the equation: ay'' +by'+cy=0
If the roots of the corresponding characteristic equation are real, show that a solution to the differential equation either is everywhere zero or else can take on the value zero at most once.
hmm I have no idea how to do this one, I think it might have to do something with repeated roots?, but im now sure, any tips/solutions to this one? thanks! :D
 A: The general solution is of the form $C_1\, \exp(r_1 x) + C_2 \, \exp(r_2x)$ where $r_1$ and $r_2$ are, in this case, the real roots of $ar^2 + br +c = 0$.
The roots are $ r_1 = -\alpha + \beta$ and $ r_2 = -\alpha - \beta $ where $\alpha = \frac{b}{2a}$ and $\beta = \sqrt{\alpha^2 -\frac{c}{a}}$.
So the general solution is
$$C_1\exp(- \alpha x)\exp(\beta x) + C_2 \exp(- \alpha x)\exp(-\beta x)$$
If $C_1 = C_2 = 0$ then the solution is everywhere zero. If $C_1 = 0$ and $C_2 \neq 0$ or if $C_1 \neq 0$ and $C_2 = 0$ then the solution is never zero because the exponential function is positive.  
Now suppose $C_1$ and $C_2$ are both non-zero. Setting the general solution to $0$ we see
$$\exp(-\alpha x)\left[C_1 \exp(\beta x) + C_2 exp(-\beta x)\right] = 0$$
Dividing both sides by $\exp(-\alpha x)$ and rearranging we get
$$\exp(2\beta x) = -\frac{C_2}{C_1}$$
which has one and only one solution for $x$ if the constants have opposite signs. In this case the solution of the differential equation is $0$ at only this single value of x.
A: *

*Consider the linear 2nd order ODE
ay′′ + by′ + cy = 0. (1) 


Since the equation is linear, with constant coefficients, and no terms not involving y (i.e. the
right hand side is 0), a logical solution to try is
y(t) = ce^(rt), (2)
where c and r are unknown constants.
(a) Why is the proposed solution (2) a logical choice for a linear equation? Hint: Think back to other ODEs we have solved. There is a class of first order ODEs that is analogous to (1).
(b) Plug the proposed solution into (1) and solve for r. (c) What three cases arise?
(c) What three cases arise?
