# Thread: y'' + by' + 4y = 0; what effect does changing b have?

1. ## y'' + by' + 4y = 0; what effect does changing b have?

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?

2. 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?
The technique is to look for a solution of the form $\displaystyle y = A e^{\lambda x}$. When you substitute this into the DE and siplify you get the auxillary equation:

$\displaystyle \lambda^2 + b \lambda + 4 = 0$.

The number of solutions for $\displaystyle \lambda$ and hence the type of solution to the DE will depend on the discriminant of this quadratic ....

1. $\displaystyle \Delta > 0$.

2. $\displaystyle \Delta = 0$.

3. $\displaystyle \Delta < 0$.

So think about the value of $\displaystyle b$ for each of these three cases ....