Question

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.

Answer 1

not even sure where to start on this one

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

method of undetermined coefficients...

Answer 6

for a second order ode....

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

it looks like it already gave you the solution

Answer 12

just apply the conditions

Answer 13

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

Answer 14

\[y=-\sin(x)+Ax+B \\ y(\frac{\pi}{2})=0 \text{ insert this into the solution }\]
you will also need to find y' given that
and then use the other condition y'(pi/2)=-2

Answer 15

you will wind up with a system of equations

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

now what is y'

Answer 20

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

Answer 21

yes otherwise you are differentiating constants on both sides and wind up with 0=0 which doesn't really help
\[y=-\sin(x)+Ax+B \\ y'=-\cos(x)+A+0 \\ y'=-\cos(x)+A\]

Answer 22

now plug in the condition

Answer 23

for y'

Answer 24

yay i found a !!!!!!

Answer 25

go back to the first equation and find B

Answer 26

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

Answer 27

because this was easier than you thought imaginable ?

Answer 28

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

Answer 29

hmmm... I might be geting pi+1

Answer 30

one sec

Answer 31

no yea your right its -2 not 2 my bad

Answer 32

okay so what now

Answer 33

\[y'(\frac{\pi}{2})=-\cos(\frac{\pi}{2})+A \\ -2=-\cos(\frac{\pi}{2})+A \\ -2 =0+A \\ -2=A \\ 0=-1+A(\frac{\pi}{2})+B \\ 0=-1-2(\frac{\pi}{2})+B \\ 0=-1-\pi+B \\ B=\pi+1\]
just wanted to make sure

Answer 34

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

Answer 35

just plug them back in

Answer 36

into the solution they gave you

Answer 37

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

Answer 38

holy crap i found it !!! its D

AAAAAHH this is ssoo cool !!!

Answer 39

lol you're funny

Answer 40

im such a pro !!!!

Answer 41

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

Answer 42

thanks for the help , i really appreciate it!

Answer 43

np