Question

Find the particular solution to y " = 3sin(x) given the general solution y = −3sin(x) + Ax + B and the initial conditions y of pi over 2 equals pi and y prime of pi over 2 equals 2.

Answer 1

https://i.gyazo.com/e84939a3da90b965eb11fe1b9563f3ee.png

Answer 2

@freckles How would I plug it in here?

Answer 3

k you will have a system
notice one of your conditions is for y'

Answer 4

so you will need to find y'

Answer 5

First take the derivative of y and plug in pi/2

Answer 6

you will use y(pi/2)=pi for y=-3sin(x)+Ax+B
and use y'(pi/2)=2 for y'=blah
blah being the derivative of -3sin(x)+Ax+B

Answer 7

Answer is A, correct?

Answer 8

\[y=-3 \sin(x)+Ax+B \\ y'=-3 \cos(x)+A \\ \text{ we have } y'(\frac{\pi}{2})=2 \\ 2=- 3 \cos(\frac{\pi}{2})+A \\ 2=-3(0)+A \\ 2=A\]
I'm getting a different A then them

Answer 9

they have -2

Answer 10

Same, I just thought it had something to with the -3

Answer 11

Even though I know that's not how algebra works.

Answer 12

\[y=-3 \sin(x)+2x+B \\ \text{ using } y(\frac{\pi}{2})=\pi \\ \text{ we have } \\ \pi=-3 \sin(\frac{\pi}{2})+2(\frac{\pi}{2})+B \\ \pi=-3(1)+\pi+B \\ 0=-3+B \\ B=3 \\ \text{ so the solution should be } y=-3 \sin(x)+2x+3\]

Answer 13

the first one would probably be the one I pick even though they are all wrong :p

Answer 14

Same. I'll even tell me teacher for some brownie points :p thanks

Answer 15

k k we all make mistakes
we shouldn't be too hard on teachers though in my opinion
teachers put up with a lot of stuff