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anonymous
 one year ago
Yet another Calculus question im having trouble with .
Find the particular solution to y " = sin(x) given the general solution y = −sin(x) + Ax + B and the initial conditions y(pi/2) = 0 and y '(pi/2) = 2.
anonymous
 one year ago
Yet another Calculus question im having trouble with . Find the particular solution to y " = sin(x) given the general solution y = −sin(x) + Ax + B and the initial conditions y(pi/2) = 0 and y '(pi/2) = 2.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not even sure where to start on this one

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you make a guess for the particular solution I know the derivative of sin(x) is cos(x) and the derivative of cos(x) is sin(x) so I might try something like \[ y_p=A sin(x)+Bcos(x)\] for the particular solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you can call A and B something else since they already being used

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm given the answer choices do you think they might help?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0method of undetermined coefficients...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0for a second order ode....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'v never been thought second order.... are they much different?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yea you will get something like Asin(x)Bcos(x)=sin(x) shich means B=0 and A=1 so A=1 so we didn't need the cos part of the particular solution

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0second order odes are easier than first order odes... that's what a ton of ode students say

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these area the answer choices by the way −sin(x) +1 + π −sin(x) − 2x + π −sin(x) − 2x − π −sin(x) − 2x + π + 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4it looks like it already gave you the solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.4just apply the conditions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean .... im not even sure what to solve for

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[y=\sin(x)+Ax+B \\ y(\frac{\pi}{2})=0 \text{ insert this into the solution }\] you will also need to find y' given that and then use the other condition y'(pi/2)=2

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you will wind up with a system of equations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you mean \[y(\frac{ \pi }{ 2 }) = 1 + Api + B\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4not exactly \[y(\frac{\pi}{2})=\sin(\frac{\pi}{2})+A(\frac{\pi}{2})+B \\ 0=1+A(\frac{\pi}{2})+B\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0x will be pi/2 and then that would be = 0 just like freckies put it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do we have to derive sin(x) + Ax + B before plugging in pi/2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yes otherwise you are differentiating constants on both sides and wind up with 0=0 which doesn't really help \[y=\sin(x)+Ax+B \\ y'=\cos(x)+A+0 \\ y'=\cos(x)+A\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4now plug in the condition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay i found a !!!!!!

freckles
 one year ago
Best ResponseYou've already chosen the best response.4go back to the first equation and find B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay ......aahhh im so exited for some reason !!!

freckles
 one year ago
Best ResponseYou've already chosen the best response.4because this was easier than you thought imaginable ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0possibly ... okay so im getting B = pi +1 is that right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4hmmm... I might be geting pi+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no yea your right its 2 not 2 my bad

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[y'(\frac{\pi}{2})=\cos(\frac{\pi}{2})+A \\ 2=\cos(\frac{\pi}{2})+A \\ 2 =0+A \\ 2=A \\ 0=1+A(\frac{\pi}{2})+B \\ 0=12(\frac{\pi}{2})+B \\ 0=1\pi+B \\ B=\pi+1\] just wanted to make sure

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well as said you already had the solution the conditions just needed to be applied so you found A and B

freckles
 one year ago
Best ResponseYou've already chosen the best response.4just plug them back in

freckles
 one year ago
Best ResponseYou've already chosen the best response.4into the solution they gave you

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[y=\sin(x)+Ax+B \\ \text{ we found } A=2 \\ \text{ and } B=\pi+1 \\ \text{ plug \in and done }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0holy crap i found it !!! its D AAAAAHH this is ssoo cool !!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay maybe nmot .... but still !!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for the help , i really appreciate it!
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