Someone help me im DYING with this (1 Viewer)

[InteGrand]

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

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

 

[hayabusaboston]

so rough solution is
q(t)=Atcos(sqrt20.t)+Btsin(sqrt20.t)
q'(t)=Acos(sqrt20.t)-(sqrt20)Atsin(sqrt20.t)+Bsin(sqrt20.t)+(sqrt20)Btcos(sqrt20.t)

with q(0)=2 and q'(0)=0 it appears to screw up.... did I do something wrong?

 

[InteGrand]

so rough solution is
q(t)=Atcos(sqrt20.t)+Btsin(sqrt20.t)
q'(t)=Acos(sqrt20.t)-(sqrt20)Atsin(sqrt20.t)+Bsin(sqrt20.t)+(sqrt20)Btcos(sqrt20.t)

with q(0)=2 and q'(0)=0 it appears to screw up.... did I do something wrong?

We don't use the initial conditions now, they're not what determine the A and B. The ODE is what determines them. We need to substitute our particular solution into the ODE and use that to solve for A and B. Then we will be able to get our general solution with two arbitrary constants in it (inside the homogenous part of the solution), and then we can use the initial conditions to get these arbitrary constants.

 

[hayabusaboston]

We don't use the initial conditions now, they're not what determine the A and B. The ODE is what determines them. We need to substitute our particular solution into the ODE and use that to solve for A and B. Then we will be able to get our general solution with two arbitrary constants in it (inside the homogenous part of the solution), and then we can use the initial conditions to get these arbitrary constants.

OH RIGHT yea

cos this is still finding that particular solution



the initial conditions are used one the particular solution equation



not the prerequisite equation BEFORE the particular solution equation

 

[hayabusaboston]

omg


A I get -50/sqrt 20 but there is no way to find a B since Bt terms disappear and B terms also disappear.

 

[hayabusaboston]

Nope all good just a minus sign mess up Lel

 

[hayabusaboston]

is d) answer

qres(t)=2cos(sqrt20.t)+(5/2)sin(sqrt20.t)-(50t/sqrt20)cos(sqrt20.t)?

 

[InteGrand]

is d) answer

qres(t)=2cos(sqrt20.t)+(5/2)sin(sqrt20.t)-(50t/sqrt20)cos(sqrt20.t)?

Well if you have something you think is the answer, you can check it by subbing it into the given ODE and also checking if it satisfies the initial conditions.

 

[akiratwang]

HAHAHAHA CALC 2 ASSIGNMENT 4 LIFE. I was stuck on Q2d) as well!

 

[hayabusaboston]

HAHAHAHA CALC 2 ASSIGNMENT 4 LIFE. I was stuck on Q2d) as well!

u doing calc 2 this sem?

 

[akiratwang]

yeh im in the 3rd lecture stream atm doing the assignment. i was getting the 100sin(2root5t)=0 as well but still not convinced why its Atcos(2root5t)+Btsin(2root5t)

 

[hayabusaboston]

yeh im in the 3rd lecture stream atm doing the assignment. i was getting the 100sin(2root5t)=0 as well but still not convinced why its Atcos(2root5t)+Btsin(2root5t)

im not convinced either. Sqrt 20 isnt same as root 20 so I dont see why u cant just use Acos+Bsin

 

[akiratwang]

ive been able to successfully solve for q(t) where w=2root5 but do we need to resolve both c1 and c2 constants? my final answer for d) was q(t)=c1*cos(2root5*t)+c2*sin(2root5*t)-5root5*t*cos(2root5*t) which i believe is right. My lecturer is the chinese one so her english isnt very good - i havent turned up to any since the limit laws HAHAHAHA

 

[hayabusaboston]

m8 I got d)


q(t)=2cos20t+5sqrt5.t.cos(sqrt20.t)

 

[hayabusaboston]

ive been able to successfully solve for q(t) where w=2root5 but do we need to resolve both c1 and c2 constants? my final answer for d) was q(t)=c1*cos(2root5*t)+c2*sin(2root5*t)-5root5*t*cos(2root5*t) which i believe is right. My lecturer is the chinese one so her english isnt very good - i havent turned up to any since the limit laws HAHAHAHA

watch the 10.05am lecture with the german dude. He writes up everything step by step on the recording so you can see whats happening, basically everyone else writes stuff on the board at the lecture so u cant get a handle on whats going on.


If u watch german dude you'll get everything you need to know, no need to attend lectures.

 

[hayabusaboston]

ive been able to successfully solve for q(t) where w=2root5 but do we need to resolve both c1 and c2 constants? my final answer for d) was q(t)=c1*cos(2root5*t)+c2*sin(2root5*t)-5root5*t*cos(2root5*t) which i believe is right. My lecturer is the chinese one so her english isnt very good - i havent turned up to any since the limit laws HAHAHAHA

yes find constants, it says solve initial value problem for resonance so use the values

 

[akiratwang]

aight so i have to then resolve for intial values wih q(0)=2 and q'(0)=0 for c1 and c2 then put it back for q(t)?

 

[akiratwang]

hmmm i seem to be getting 2cos(2root5*t)+(5/2)sin(2root5*t)

 

[hayabusaboston]

pls halp

in which direction in xy plane is directional derivative at (1,1) equal to zero of f(x,y)=(x^2-y^2)/(x^2+y^2)