Someone help me im DYING with this (1 Viewer)

InteGrand · May 9, 2017

hayabusaboston said:
so if sin(wt) is equal to the general solution values of sin and cos?

characteristic equation results in 20i,-20i so y(t)=Acos20t+Bsin20t, if times by t will be equal with sinwt so

20 is it? later questions say "pick a value where resonance occurs" isnt there only the one?

Yeah there's only one positive value, but you could also use -20 and that'd still give resonance. Usually omega is thought of as positive I guess. Using a negative omega here just makes the RHS of your ODE become negative sine, since sine is odd. This means your ''particular solution'' becomes negative of the old one.

hayabusaboston · May 9, 2017

InteGrand said:
Yeah there's only one positive value, but you could also use -20 and that'd still give resonance. Usually omega is thought of as positive I guess. Using a negative omega here just makes the RHS of your ODE become negative sine, since sine is odd. This means your ''particular solution'' becomes negative of the old one.

for c)do you just get rid of w and say sint then do initial value problem?

for d) do you just solve the initial problem with w=20?

InteGrand · May 9, 2017

hayabusaboston said:
for c)do you just get rid of w and say sint then do initial value problem?

for d) do you just solve the initial problem with w=20?

$\noindent For c), you'd do it by solving it with a general $\omega$ where $\omega \neq \omega_{0}$, where $\omega_{0}$ is the natural frequency, which is the \textbf{square root} of $20$. I.e. solve $q'' + 20 q = 100 \sin \omega t$, where $\omega \neq \sqrt{20}$ (with the given initial values). For d), you'd solve the IVP with $\omega = \sqrt{20}$. As you'll find, the form of the particular solution is different in each case.$

hayabusaboston · May 9, 2017

InteGrand said:
$\noindent For c), you'd do it by solving it with a general $\omega$ where $\omega \neq \omega_{0}$, where $\omega_{0}$ is the natural frequency, which is the \textbf{square root} of $20$. I.e. solve $q'' + 20 q = 100 \sin \omega t$, where $\omega \neq \sqrt{20}$ (with the given initial values). For d), you'd solve the IVP with $\omega = \sqrt{20}$. As you'll find, the form of the particular solution is different in each case.$

wot how do u do that? solve w not equal to w0?

InteGrand · May 9, 2017

hayabusaboston said:
wot how do u do that? solve w not equal to w0?

$\noindent The ODE to solve is $q'' + 20 q = 100\sin \omega t$, where $\omega \neq \pm \sqrt{20}$.$

InteGrand · May 9, 2017

hayabusaboston · May 9, 2017

InteGrand said:
$\noindent The ODE to solve is $q'' + 20 q = 100\sin \omega t$, where $\omega \neq \pm \sqrt{20}$.$

but how do u solve like that?

General solution is y(t)=acos20t+bsin20t,

then for particular solution what do u do? make an Atcosw+Btsinw? Where go from there?

well u have y=Atcosw+Btsinw
y'=-Atsinw+Btcosw
y''=-Atcosw-Btsinw

edit: messed up woah

InteGrand · May 9, 2017

hayabusaboston said:
but how do u solve like that?

General solution is y(t)=acos20t+bsin20t,

then for particular solution what do u do? make an Atcosw+Btsinw? Where go from there?

$\noindent In general we'd try a particular solution of the form $A \cos \omega t + B\sin \omega t$ for the given right-hand side of the ODE, but since the LHS of the ODE is what it is, it suffices to just try $A \sin \omega t$ (if you add a cosine term in, it'll still work, you'll just find the coefficient for it is $0$).$

$\noindent Substituting $A \sin \omega t$ into the ODE, we have$

$\begin{align*}\left(A \sin \omega t\right)'' + 20 \cdot A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow -\omega^{2} A \sin \omega t + 20A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow \left(20 - \omega^{2}\right)A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow \left(20 - \omega^{2}\right)A &= 100.\end{align*}$

$\noindent Since $\omega^{2} \neq 20$ for this case (non-resonance), we have $A = \frac{100}{20 - \omega^{2}}$. So a particular solution is $q_{\text{part.}}(t) = A\sin \omega t = \frac{100}{20 - \omega^{2}} \sin \omega t$. You should now be able to write down the general solution (involving two arbitrary constants). You can then find these constants using the initial conditions.$

hayabusaboston · May 9, 2017

InteGrand said:
$\noindent In general we'd try a particular solution of the form $A \cos \omega t + B\sin \omega t$ for the given right-hand side of the ODE, but since the LHS of the ODE is what it is, it suffices to just try $A \sin \omega t$ (if you add a cosine term in, it'll still work, you'll just find the coefficient for it is $0$).$

$\noindent Substituting $A \sin \omega t$ into the ODE, we have$

$\begin{align*}\left(A \sin \omega t\right)'' + 20 \cdot A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow -\omega^{2} A \sin \omega t + 20A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow \left(20 - \omega^{2}\right)A \sin \omega t &= 100 \sin \omega t \\ \Rightarrow \left(20 - \omega^{2}\right)A &= 100.\end{align*}$

$\noindent Since $\omega^{2} \neq 20$ for this case (non-resonance), we have $A = \frac{100}{20 - \omega^{2}}$. So a particular solution is $q_{\text{part.}}(t) = A\sin \omega t = \frac{100}{20 - \omega^{2}} \sin \omega t$. You should now be able to write down the general solution (involving two arbitrary constants). You can then find these constants using the initial conditions.$

Im a retard. That was easy. I am having such difficulty braining anything today jesus.

hayabusaboston · May 9, 2017

Why such trouble concentrating :'(

hayabusaboston · May 9, 2017

my brain is sickeningly slow today.

Initially there is no current in the circuit and the initial capacitor charge is 2 coulombs. How is this represented in q(t) equations?

im doing c and have

general solution non resonance:

q(t)=acos20t+bsin20t+(100/(20-w^2))sinwt -->let 100/(20-w^2)=M
q'(t)=-20asin20t+20bcos20t+M'
q''(t)=-400acos20t-400bsin20t-M''

hasdasdasfsdfsdfsdfsdf

help PLOX how 2 solve the initial value thing its getting huge, am I doing it wrong?

InteGrand · May 9, 2017

hayabusaboston said:
my brain is sickeningly slow today.

Initially there is no current in the circuit and the initial capacitor charge is 2 coulombs. How is this represented in q(t) equations?

im doing c and have

general solution non resonance:

q(t)=acos20t+bsin20t+(100/(20-w^2))sinwt -->let 100/(20-w^2)=M
q'(t)=-20asin20t+20bcos20t+M'
q''(t)=-400acos20t-400bsin20t-M''

hasdasdasfsdfsdfsdfsdf

help PLOX how 2 solve the initial value thing its getting huge, am I doing it wrong?

Current is rate of flow of charge, so initially 0 current means q'(0) = 0. Also, initially 2 coulombs of charge means q(0) = 2.

hayabusaboston · May 9, 2017

InteGrand said:
Current is rate of flow of charge, so initially 0 current means q'(0) = 0. Also, initially 2 coulombs of charge means q(0) = 2.

Yea I realised that a minute after I posted XD

the initial value problem for c) is getting huge... I dont know if im doing it right. You're supposed to get values for everything except the w yea? since w represents all non resonance values.

so ur finding a and b from original thing.

Do u use 100sinwt or the 100/(20-w^2) sinwt

100/(20-w^2) turned so contorted I decided to stop and ask if it should be that way.

hayabusaboston · May 9, 2017

also for d) do u just use original 100sinwt but say w=root 20?

hayabusaboston · May 9, 2017

I dont see whats wrong with this but its so huge it feels wrong

I got for c)

q(t)=2(20-w^2)cos20t+100sinwtcos20t-5wcoswtsin20t+100sinwt

all over 20-w^2

is dat right?

I cant even figure out how to simplify im so tired atm

InteGrand · May 10, 2017

hayabusaboston said:
Yea I realised that a minute after I posted XD

the initial value problem for c) is getting huge... I dont know if im doing it right. You're supposed to get values for everything except the w yea? since w represents all non resonance values.

so ur finding a and b from original thing.

Do u use 100sinwt or the 100/(20-w^2) sinwt

100/(20-w^2) turned so contorted I decided to stop and ask if it should be that way.

hayabusaboston said:
I dont see whats wrong with this but its so huge it feels wrong

I got for c)

q(t)=2(20-w^2)cos20t+100sinwtcos20t-5wcoswtsin20t+100sinwt

all over 20-w^2

is dat right?

I cant even figure out how to simplify im so tired atm

$\noindent We found the general solution to be$

$q(t) = c_{1} \cos (\omega_{0} t) + c_{2}\sin (\omega_{0} t) + \frac{100}{20 - \omega^{2}}\sin (\omega t),$

$\noindent where $\omega_{0} = \sqrt{20}$ and $\omega^{2} \neq 20$. We are given the initial conditions $q(0) = 2$ and $q'(0) = 0$. Substitute these in to find $c_{1}$ and $c_{2}$. So we have$

$\begin{align*}2 &= q(0)\\ &= c_{1} \cos (\omega_{0} \times 0) + c_{2}\sin (\omega_{0} \times 0) + \frac{100}{20 - \omega^{2}}\sin (\omega \times 0) \\ &= c_{1} + 0 + 0 \quad (\text{since }\cos 0 = 1 \text{ and }\sin 0 = 0) \\ &\Rightarrow \boxed{c_{1} = 2}.\end{align*}$

$\noindent Note that$

$q'(t) = -c_{1}\omega_{0} \sin (\omega_{0} t) + c_{2}\omega_{0}\cos (\omega_{0} t) + \frac{100\omega}{20 - \omega^{2}}\cos (\omega t).$

$\noindent Hence the initial condition $q'(0) = 0$ gives us$

$\begin{align*}0 &= q'(0) \\ &= -c_{1}\omega_{0} \sin (\omega_{0} \times 0) + c_{2}\omega_{0}\cos (\omega_{0} \times 0) + \frac{100\omega}{20 - \omega^{2}}\cos (\omega \times 0)\\ \Rightarrow 0 &= c_{2}\omega_{0} + \frac{100\omega}{20 - \omega^{2}}. \end{align*}$

$$\noindent Solving for $c_{2}$ gives $\boxed{c_{2} = - \frac{100\omega}{\omega_{0}\left(20 - \omega^{2}\right)}}$.$

$\noindent Hence the solution to the IVP is$

$\boxed{q(t) = 2 \cos (\omega_{0} t) - \frac{100\omega}{\omega_{0}\left(20 - \omega^{2}\right)}\sin (\omega_{0} t) + \frac{100}{20 - \omega^{2}}\sin (\omega t)},$

$\noindent where $\omega_{0} = \sqrt{20}$ and $\omega^{2}\neq 20$.$

hayabusaboston · May 10, 2017

InteGrand said:
$\noindent We found the general solution to be$

$q(t) = c_{1} \cos (\omega_{0} t) + c_{2}\sin (\omega_{0} t) + \frac{100}{20 - \omega^{2}}\sin (\omega t),$

$\noindent where $\omega_{0} = \sqrt{20}$ and $\omega^{2} \neq 20$. We are given the initial conditions $q(0) = 2$ and $q'(0) = 0$. Substitute these in to find $c_{1}$ and $c_{2}$. So we have$

$\begin{align*}2 &= q(0)\\ &= c_{1} \cos (\omega_{0} \times 0) + c_{2}\sin (\omega_{0} \times 0) + \frac{100}{20 - \omega^{2}}\sin (\omega \times 0) \\ &= c_{1} + 0 + 0 \quad (\text{since }\cos 0 = 1 \text{ and }\sin 0 = 0) \\ &\Rightarrow \boxed{c_{1} = 2}.\end{align*}$

$\noindent Note that$

$q'(t) = -c_{1}\omega_{0} \sin (\omega_{0} t) + c_{2}\omega_{0}\cos (\omega_{0} t) + \frac{100\omega}{20 - \omega^{2}}\cos (\omega t).$

$\noindent Hence the initial condition $q'(0) = 0$ gives us$

$\begin{align*}0 &= q'(0) \\ &= -c_{1}\omega_{0} \sin (\omega_{0} \times 0) + c_{2}\omega_{0}\cos (\omega_{0} \times 0) + \frac{100\omega}{20 - \omega^{2}}\cos (\omega \times 0)\\ \Rightarrow 0 &= c_{2}\omega_{0} + \frac{100\omega}{20 - \omega^{2}}. \end{align*}$

$$\noindent Solving for $c_{2}$ gives $\boxed{c_{2} = - \frac{100\omega}{\omega_{0}\left(20 - \omega^{2}\right)}}$.$

$\noindent Hence the solution to the IVP is$

$\boxed{q(t) = 2 \cos (\omega_{0} t) - \frac{100\omega}{\omega_{0}\left(20 - \omega^{2}\right)}\sin (\omega_{0} t) + \frac{100}{20 - \omega^{2}}\sin (\omega t)},$

$\noindent where $\omega_{0} = \sqrt{20}$ and $\omega^{2}\neq 20$.$

O jeez man.... my "constants" search was huge because I didnt realise sin0 is 0. Lmfao.

for d) you just solve q''+20q=100sin((sqrt20)t) right?>

I appreciate your help SO MUCH <3

InteGrand · May 10, 2017

hayabusaboston said:
O jeez man.... my "constants" search was huge because I didnt realise sin0 is 0. Lmfao.

for d) you just solve q''+20q=100sin((sqrt20)t) right?>

I appreciate your help SO MUCH <3

Yes, that's what you solve for part (d).

hayabusaboston · May 10, 2017

InteGrand said:
Yes, that's what you solve for part (d).

wot I get -20Asin(sqrt20.t)+20Asin(sqrt20.t) matches to 100sin(sqrt20t) when coefficient matching.

How do u find A...? equals zero mann

Someone help me im DYING with this (1 Viewer)

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