## 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.

1. anonymous

not even sure where to start on this one

2. 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

3. freckles

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

4. anonymous

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

5. UsukiDoll

method of undetermined coefficients...

6. UsukiDoll

for a second order ode....

7. anonymous

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

8. 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

9. UsukiDoll

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

10. 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

11. freckles

it looks like it already gave you the solution

12. freckles

just apply the conditions

13. anonymous

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

14. 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

15. freckles

you will wind up with a system of equations

16. anonymous

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

17. freckles

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

18. UsukiDoll

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

19. freckles

now what is y'

20. anonymous

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

21. freckles

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

22. freckles

now plug in the condition

23. freckles

for y'

24. anonymous

yay i found a !!!!!!

25. freckles

go back to the first equation and find B

26. anonymous

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

27. freckles

because this was easier than you thought imaginable ?

28. anonymous

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

29. freckles

hmmm... I might be geting pi+1

30. freckles

one sec

31. anonymous

32. anonymous

okay so what now

33. 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

34. freckles

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

35. freckles

just plug them back in

36. freckles

into the solution they gave you

37. freckles

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

38. anonymous

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

39. freckles

lol you're funny

40. anonymous

im such a pro !!!!

41. anonymous

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

42. anonymous

thanks for the help , i really appreciate it!

43. freckles

np