# Show that `(e^(rx),xe^(rx))` is not equal to 0 for any value of r. Conclude that if `e^(rx)` and `xe^(rx)` are solutions of ` ay'' + by' + cy=0` , they are a fundamental set of solutions.

*print*Print*list*Cite

### 1 Answer

Alright, we'll start off the problem by sketching a proof that `e^(rx)` can never be 0 given changes in `r`:

First, we'll note that we can generalize the negative case to the positive case simlply by reflection over the x-axis. This means that if we can prove or disprove that `e^(rx) = 0` for the case where `r>0`, then that result would imply the same for when `r<0`. If `r=0` , then the function becomes a constant value, 1, which (of course) can never be zero. So, we'll just concentrate on the case where `r>0`.

We'll start by noting that, based on the Taylor Series representation of `e^(rx)` that the function is monotonically increasing. We'll also take note of the fact that it is a continuous function. Finally, based on its y-intercept of (0,1), we know that the function is postive given `x>0`.

Because it is monotonically increasing, given a point (`x_1`,`e^(rx_1))` we know there will be two more points `(x_2,e^(rx_2))` and `(x_3, e^(rx_3))`where `x_2<x_1<x_3` and `e^(rx_2) < e^(rx_1) < e^(rx_3)`. So, if we had a situation where `e^(rx_1) = 0`, then we would necessarily have a point `(x_2, e^(rx_2)) ` where `e^(rx_2)<0`. Because we already found that `e^(rx)>0` if `x>0` , we know that our only shot of `e^(rx_2) < 0` is if `x_2<0`. However, if `x_2<0` let's have a corresponding `x_(2+) = -x_2`. Our function would then become:

`e^(rx_2) = e^(-rx_(2+)) = 1/e^(rx_(2+))`

However, we know that `e^(rx_(2+))` must be positive because `x_(2+)` must be positive. Therefore, `e^(rx_2)` must be positive as well! This contradicts the possibility of there being a point at which the exponential function is equal to zero.

**Therefore, there is no point at which the exponential function can be zero.**

To prove the same for `xe^(rx)` all you would need to do is have the same proof, but have your points be in the form of `(r, xe^(rx))`. In other words, you treat `r` as the independent variable and `x` as the constant.

Now, moving on to the differential equation. You only get the solution set `e^(rx), xe^(rx)` if the roots of your characteristic polynomial are both `r`.

Let's first point out that `e^(rx)` and `xe^(rx)` are linearly independent, as they are not related by a common factor. Because they are two linearly independent solutions of a second-order differential equation, according to the problem, and because a basis of a second-order differential equation is formed by two linearly independent solutions, **they necessarily form a basis (fundamental set of solutions).**

**Side Note**: These two solutions both result from the characteristic polynomial `lambda^2 + b/a lambda + c/a = 0` giving us a perfect square trinomial (which we'll let the double root be "r"). The method by which you show that `xe^(rx)` is a solution to this specific situation given that `e^(rx)` is through a method called "reduction of order." This method, in general, is based on having found a `y_1` which is a homogeneous solution and letting the second solution, `y_2`, be:

`y _2 = uy_1`

where `u` is just another function. When you solve the perfect-square trinomial second-order D.E. with this method, you find that `u = x` giving your `y_2 = xe^(rx)`.

**Sources:**