Basic 2nd Order ODE Question: aux. eqn ar^2 + br + c = 0

[BlazingFire]

Show that if the auxiliary equation \(\displaystyle ar^2 + br + c = 0\) has a repeated real-value root \(\displaystyle r_1\), then \(\displaystyle y_2(t) = te^{r_1(t)}\) is a solution to the ODE \(\displaystyle ay'' + by' + cy = 0\). Can anyone help me do this without using the Wronskian?

 

[royhaas]

Re: Basic 2nd Order ODE Question

Did you substitute \(\displaystyle y_2\) into the ODE? Note that we must have \(\displaystyle r_1 = \frac{-b}{2a}\).

 

[BlazingFire]

I took \(\displaystyle y_2(t) = te^{r_1(t)}\) and its two derivatives with respect to t and plugged them into \(\displaystyle ay'' + by' + cy = 0\) (BTW, if you spot any errors on my part, please let me know right away).

First off, I'll start with the first and second derivatives of \(\displaystyle y_2(t) = te^{r_1(t)}\) as I calculated them. I'm pretty confident that these should be right, but anyone can double-check if they'd like (especially with the second derivative):

\(\displaystyle y'_2(t) = r_1te^{r_1(t)} + e^{r_1(t)}\)
\(\displaystyle y''_2(t) = {r_1}^2te^{r_1(t)} + r_1 e^{r_1(t)} + r_1 e^{r_1(t)} = {r_1}^2te^{r_1(t)} + 2r_1 e^{r_1(t)}\)

Now plugging this into \(\displaystyle ay'' + by' + cy = 0\) gives me:

\(\displaystyle a({r_1}^2te^{r_1(t)} + 2r_1 e^{r_1(t)}) + b(r_1te^{r_1(t)} + e^{r_1(t)}) + cte^{r_1(t)} = 0\)

Do I really have to simplify this big mess or what? Will that be necessary to complete this problem?

I'm not sure I see this getting anywhere.

Seriously, help me out here. I'm just going by what you said, and if anyone else has anything substantive to add...go for it, and explain step-by-step.

 

[royhaas]

Check your final expression. Looks like you used the second derivative twice. Use \(\displaystyle c=\frac{b^2}{4a}\) and \(\displaystyle r_1=\frac{-b}{2a}\) in simplifying.

 

[Deleted member 4993]

I took \(\displaystyle y_2(t) = te^{r_1(t)}\) and its two derivatives with respect to t and plugged them into \(\displaystyle ay'' + by' + cy = 0\) (BTW, if you spot any errors on my part, please let me know right away).

First off, I'll start with the first and second derivatives of \(\displaystyle y_2(t) = te^{r_1(t)}\) as I calculated them. I'm pretty confident that these should be right, but anyone can double-check if they'd like (especially with the second derivative):

\(\displaystyle y'_2(t) = r_1te^{r_1(t)} + e^{r_1(t)}\)
\(\displaystyle y''_2(t) = {r_1}^2te^{r_1(t)} + r_1 e^{r_1(t)} + r_1 e^{r_1(t)} = {r_1}^2te^{r_1(t)} + 2r_1 e^{r_1(t)}\)

Now plugging this into \(\displaystyle ay'' + by' + cy = 0\) gives me:

\(\displaystyle a({r_1}^2te^{r_1(t)} + 2r_1 e^{r_1(t)}) + b(r_1te^{r_1(t)} + e^{r_1(t)}) + cte^{r_1(t)} = 0\) .... corrected

\(\displaystyle te^{r_1(t)} \cdot (a{r_1}^2 + br_1 +c) \, + \, e^{r_1(t)} \cdot (2ar_1 + b) = 0\)

Do I really have to simplify this big mess or what? ..... Yes

Will that be necessary to complete this problem? .... Yes - as far as I know.

I'm not sure I see this getting anywhere.

Seriously, help me out here. I'm just going by what you said, and if anyone else has anything substantive to add...go for it, and explain step-by-step.