# Second order confusion

1. Feb 8, 2017

### Taylor_1989

1. The problem statement, all variables and given/known data
Hi guys, I am having trouble with ex 6.1 and 6.2. I am listed my ans below, and shown working. Could someone please advise. First I am not sure why I have to start with $$x=Ae^{imt}$$. Why cannot start with
$$Ae^{px}$$?

2. Relevant equations

3. The attempt at a solution
i) $$Aux: -m^2+irm+r^2=0$$
$$\Delta=(ir)^2-4(-1)(r^2)=3r^2$$
$$\sqrt(\Delta)=r\sqrt(3)$$
$$m=\frac{ir\pm r\sqrt(3)}{2}$$

$$x=e^{\frac{-ir}{2}}(Ae^{\frac{ir\sqrt3}{2}}+Be^{\frac{-ir\sqrt3}{2}})$$

ii) For this I followed the same process but came up with, which not enitrly sure if that satisfys the general solution.

$$x=Ae^{\frac{-ir}{2}}$$

iii) Once again same process and I got

$$x=e^{\frac{ir}{2}}(Ae^{\frac{ir\sqrt5}{2}}+Be^{\frac{-ir\sqrt5}{2}})$$

6.2 why do I have to concern myself with the solution, of
$$y=Axe^{imt}$$
when I am finding the P.I?

2. Feb 8, 2017

### Staff: Mentor

You skipped too many steps. The first two terms above are OK, but the third term is wrong.
I don't know. The image you posted doesn't include the differential equation.

3. Feb 8, 2017

### Taylor_1989

@Mark44 when you say the third is wrong which are in ref to? the quadtratic? The question say I have to use

$$Ae^{imt}$$

are you saying I should replace im with p?

4. Feb 8, 2017

### Taylor_1989

My confusion come from the question it self. Am I looking for a general soulution of the forum, x=Ae^imx or am I take the 1st and 2nd differntial and and subbing in

5. Feb 8, 2017

### Staff: Mentor

You're looking for a solution of the form $x = Ae^{imt}$, not $Ae^{imx}$. It's probably a typo, but you don't want to assume that $x = Ae^{imx}$ is a solution, with x as both the dependent and independent variables.
Find x'(t) and x''(t), and substitute into the differential equation. It looks like you were doing this to get the auxiliary equation you showed, but the $r^2$ term in your quadratic equation is wrong.

As for your question, "why can't I start with $Ae^{px}$?" the problem statement says to use $x = Ae^{imt}$ and not otherwise.

Last edited: Feb 8, 2017
6. Feb 8, 2017

### Taylor_1989

@Mark44 how is the r^2 value wrong as I have to sub it in for labda

my working are as follows:

$$\lambda=r^2$$

$$x''=-A(m^2)e^{imt}, x'= A(im)e^{imt}, x= Ae^{imt}$$
subbing the values in and factoring Ae^imt out of the equation I get a quadtratic in m

$$-m^2+imr+r^2=0$$

Last edited: Feb 8, 2017
7. Feb 8, 2017

### Staff: Mentor

I missed that (case i)bit in your posted image. Sorry to have caused some confusion.

8. Feb 8, 2017

### Staff: Mentor

Or $m^2 -imr - r^2 = 0$
Now solve for m to get your solution $x = Ae^{imt}$.

9. Feb 8, 2017

### Taylor_1989

my general solution would be:

$$x=Ae^{\frac{ir\pm \sqrt3 t}{2}}$$

10. Feb 8, 2017

### Taylor_1989

no worries the print screen is not great

11. Feb 8, 2017

### Staff: Mentor

It was clear enough -- I just didn't read far enough.
This is close.

From your solution for m, $m_1 = \frac r 2(\sqrt 3 + i), m_2 = \frac r 2(-\sqrt 3 + i)$
So, $im_1 = \frac r 2(-1 + \sqrt 3 i)$ and $im_2 = \frac r 2(-1 - \sqrt 3 i)$
$x_1(t) = Ae^{im_1t}$ and $x_2(t) = Ae^{im_2t}$
It would be a good idea to verify that both are solutions to your diff. equation.

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