Finding the General Solution for Second Order Confusion

[Taylor_1989]

Homework Statement

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
$$Ae^{px}$$?

Homework Equations

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.[/B]
$$x=Ae^{\frac{-ir}{2}}$$

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?

 

[Mark44]

Homework Statement

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
$$Ae^{px}$$?
View attachment 112754

Homework Equations

The Attempt at a Solution

i) $$Aux: -m^2+irm+r^2=0$$

You skipped too many steps. The first two terms above are OK, but the third term is wrong.

$$\Delta=(ir)^2-4(-1)(r^2)=3r^2$$
$$\sqrt(\Delta)=r\sqrt(3)$$
$$m=\frac{ir\pm r\sqrt(3)}{2}$$

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

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?

I don't know. The image you posted doesn't include the differential equation.

 

[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 I am with p?

 

[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 differential and and subbing in

 

[Mark44]

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 differential and and subbing in

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.

 

[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

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

 

[Mark44]

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

 

[Mark44]

@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

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

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

 

[Taylor_1989]

my general solution would be:

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

 

[Taylor_1989]

no worries the print screen is not great

 

[Mark44]

no worries the print screen is not great

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

my general solution would be:

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

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.