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

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- anonymous

not even sure where to start on this one

- 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

you can call A and B something else since they already being used

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## More answers

- anonymous

I'm given the answer choices do you think they might help?

- UsukiDoll

method of undetermined coefficients...

- UsukiDoll

for a second order ode....

- anonymous

i'v never been thought second order.... are they much different?

- 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

second order odes are easier than first order odes... that's what a ton of ode students say

- 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

it looks like it already gave you the solution

- freckles

just apply the conditions

- anonymous

what do you mean .... im not even sure what to solve for

- 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

you will wind up with a system of equations

- anonymous

so you mean
\[y(\frac{ \pi }{ 2 }) = -1 + Api + B\]

- freckles

not exactly
\[y(\frac{\pi}{2})=-\sin(\frac{\pi}{2})+A(\frac{\pi}{2})+B \\ 0=-1+A(\frac{\pi}{2})+B\]

- UsukiDoll

x will be pi/2
and then that would be = 0 just like freckies put it

- freckles

now what is y'

- anonymous

do we have to derive -sin(x) + Ax + B before plugging in pi/2?

- 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

now plug in the condition

- freckles

for y'

- anonymous

yay i found a !!!!!!

- freckles

go back to the first equation and find B

- anonymous

okay ......aahhh im so exited for some reason !!!

- freckles

because this was easier than you thought imaginable ?

- anonymous

possibly ... okay so im getting B = -pi +1 is that right?

- freckles

hmmm... I might be geting pi+1

- freckles

one sec

- anonymous

no yea your right its -2 not 2 my bad

- anonymous

okay so what now

- 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

well as said you already had the solution
the conditions just needed to be applied
so you found A and B

- freckles

just plug them back in

- freckles

into the solution they gave you

- freckles

\[y=-\sin(x)+Ax+B \\ \text{ we found } A=-2 \\ \text{ and } B=\pi+1 \\ \text{ plug \in and done }\]

- anonymous

holy crap i found it !!! its D
AAAAAHH this is ssoo cool !!!

- freckles

lol you're funny

- anonymous

im such a pro !!!!

- anonymous

okay maybe nmot .... but still !!!

- anonymous

thanks for the help , i really appreciate it!

- freckles

np

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