anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
not even sure where to start on this one
freckles
  • freckles
you 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
  • freckles
you can call A and B something else since they already being used

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More answers

anonymous
  • anonymous
I'm given the answer choices do you think they might help?
UsukiDoll
  • UsukiDoll
method of undetermined coefficients...
UsukiDoll
  • UsukiDoll
for a second order ode....
anonymous
  • anonymous
i'v never been thought second order.... are they much different?
freckles
  • freckles
yea 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
  • UsukiDoll
second order odes are easier than first order odes... that's what a ton of ode students say
anonymous
  • anonymous
these 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
  • freckles
it looks like it already gave you the solution
freckles
  • freckles
just apply the conditions
anonymous
  • anonymous
what do you mean .... im not even sure what to solve for
freckles
  • freckles
\[y=-\sin(x)+Ax+B \\ y(\frac{\pi}{2})=0 \text{ insert this into the solution }\] you will also need to find y' given that and then use the other condition y'(pi/2)=-2
freckles
  • freckles
you will wind up with a system of equations
anonymous
  • anonymous
so you mean \[y(\frac{ \pi }{ 2 }) = -1 + Api + B\]
freckles
  • freckles
not exactly \[y(\frac{\pi}{2})=-\sin(\frac{\pi}{2})+A(\frac{\pi}{2})+B \\ 0=-1+A(\frac{\pi}{2})+B\]
UsukiDoll
  • UsukiDoll
x will be pi/2 and then that would be = 0 just like freckies put it
freckles
  • freckles
now what is y'
anonymous
  • anonymous
do we have to derive -sin(x) + Ax + B before plugging in pi/2?
freckles
  • freckles
yes 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
  • freckles
now plug in the condition
freckles
  • freckles
for y'
anonymous
  • anonymous
yay i found a !!!!!!
freckles
  • freckles
go back to the first equation and find B
anonymous
  • anonymous
okay ......aahhh im so exited for some reason !!!
freckles
  • freckles
because this was easier than you thought imaginable ?
anonymous
  • anonymous
possibly ... okay so im getting B = -pi +1 is that right?
freckles
  • freckles
hmmm... I might be geting pi+1
freckles
  • freckles
one sec
anonymous
  • anonymous
no yea your right its -2 not 2 my bad
anonymous
  • anonymous
okay so what now
freckles
  • freckles
\[y'(\frac{\pi}{2})=-\cos(\frac{\pi}{2})+A \\ -2=-\cos(\frac{\pi}{2})+A \\ -2 =0+A \\ -2=A \\ 0=-1+A(\frac{\pi}{2})+B \\ 0=-1-2(\frac{\pi}{2})+B \\ 0=-1-\pi+B \\ B=\pi+1\] just wanted to make sure
freckles
  • freckles
well as said you already had the solution the conditions just needed to be applied so you found A and B
freckles
  • freckles
just plug them back in
freckles
  • freckles
into the solution they gave you
freckles
  • freckles
\[y=-\sin(x)+Ax+B \\ \text{ we found } A=-2 \\ \text{ and } B=\pi+1 \\ \text{ plug \in and done }\]
anonymous
  • anonymous
holy crap i found it !!! its D AAAAAHH this is ssoo cool !!!
freckles
  • freckles
lol you're funny
anonymous
  • anonymous
im such a pro !!!!
anonymous
  • anonymous
okay maybe nmot .... but still !!!
anonymous
  • anonymous
thanks for the help , i really appreciate it!
freckles
  • freckles
np

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