Originally Posted by

**BlazingFire** Here's the full text of the question:

**To see the effect of changing the parameter b in the initial value problem:**

**$\displaystyle y'' + by' + 4y = 0; y(0) = 1, y'(0) = 0$**

**Solve for $\displaystyle b = 5$, $\displaystyle b = 4$ and $\displaystyle b = 2$ and sketch the solutions.**

So basically, in short, we have the following IVPs:

1. $\displaystyle y'' + 5y' + 4y = 0; y(0) = 1, y'(0) = 0$

2. $\displaystyle y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0$

3. $\displaystyle y'' + 2y' + 4y = 0; y(0) = 1, y'(0) = 0$

I already worked out all of them. For #1, I got $\displaystyle y = -\frac{1}{3}e^{4x} + \frac{4}{3}e^{-x}$. For #2, I wound up with $\displaystyle y = e^{-2x} + 2xe^{-2x}$. And lastly, for #3, I got $\displaystyle y = e^{-x}cos(\sqrt{3}x) + \frac{\sqrt{3}}{3}e^{-x}sin(\sqrt{3}x)$. I graphed all of these functions on my calculator to look for any similarities (I swear I didn't make any mistakes in doing so; I was extremely careful with the notation), but couldn't find any.

My question is this: what effect does changing the coefficient of the first derivative have on the solution of a second-order, linear, homogeneous ODE with constant coefficients, especially in my particular case?