## amistre64 3 years ago $y''-4y=(x^2-3)sin(2x)$ solve the diffyQ; oy what a nightmare ...

1. Lukecrayonz

I'd help you if I could, but when you solve this I posted one for you to help me with.

2. amistre64

$y=y_h+y_p$ $y_h=c_1 e^{2x}+c_2 e^{-2x}$ its that yp thats making me loose my religion :)

3. Calculus102

Wooooooow. Just wondering, where did you get this problem from?

4. Lukecrayonz

Where the hell did c come into play?

5. amistre64

i call it #14 form section 4.3 ... from HELL!!! lol

6. Hermeezey

@Calculus102 It's differential equations. It's like a class after Calc III

7. Calculus102

Calc 3??? Like multivariate calculus?

8. Lukecrayonz

I'm done here, I am zero help.

9. Lukecrayonz

I'm in PRE calculus. This is madness.

10. amistre64

$y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)$

11. amistre64

calciii is cake compared to having to deal with this tripe ....

12. bahrom7893

Bernoulli's method, anyone?

13. krypton

this is very simple ,chain rule,product rule and some pellet mixed together

14. amistre64

burnoodle aint gonna help on this one

15. bahrom7893

wait nevermind... bernoulli was if u had y^n lol not nth derivative.. i suck at math

16. amistre64

its keeping track of the mindnumbingness till the end ... i dont think i have a happy personality left in me after this lol

17. bahrom7893

I took diff eq's.. use the char equation? r^2-4r = 0 r(r-4)=0 r = 0, r = 4

18. bahrom7893

but what about the right? i forgot.. and i only took this like a year ago.

19. amistre64

yeah the homogenous parts a breeze done did that

20. amistre64

the right is one of those "make a good guess" type deals

21. amistre64

heres my guess $y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)$

22. bahrom7893

ohh one of the rules? anyway i gotta eat dinner, mom's yelling at me lol.

23. amistre64

$y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)$ $y' _p=(2ax+b)sin(2x)+(2ax^2+2bx+2c)cos(2x)$$\hspace{5em} +(2fx+g)cos(2x)+(-2fx^2-2gx-2h)sin(2x)$ $y'' _p=(2a)sin(2x)+(4ax+2b)cos(2x)$$\hspace{3em}+(4ax+2b)cos(2x)+(-4ax^2-4bx-4c)sin(2x)$$\hspace{6em} +(2f)cos(2x)+(-4fx-2g)sin(2x)$$\hspace{9em}+(-4fx-2g)sin(2x)+(-4fx^2-4gx-4h)cos(2x)$

24. amistre64

$(-4ax^2-4bx-4c)sin(2x)$$(2a-4ax^2-4bx-4c-4fx-2g-4fx-2g)sin(2x)$ $(-4fx^2-4gx-4h)cos(2x)$$(4ax+2b+4qx+2b+2f-4fx^2-4gx-4h)cos(2x)$ $(-8ax^2-8bx-8c+2a-8fx-4)sin(2x)=(x^2-3)sin(2x)$$a=-\frac{1}{8}$ $(-8bx-8c-\frac{1}{4}-8fx-4)=(-3)$ ugh ... messed it up again $(-8fx^2-8gx-8h+4b+8ax+4b+2f)cos(2x)=0cos(2x)$

25. krypton

LOL...THIS CAN BE pelletY

26. bahrom7893

satellite, where does your math knowledge come from?

27. bahrom7893

Wait LOOOLL, u're amistre!!! Doesn't matter man, where does your knowledge come from amistre?

28. bahrom7893

You guys are epic! Like seriously! I'm very impressed, look up to you guys :))!

29. amistre64

$(-8ax^2-8bx-8c+2a-8fx-4g)=(x^2-3)$ $(-8a)x^2+(-8b-8f)x+(2a-8c-4g)=x^2+0x-3$ i gots no math knowledge ;)

30. amistre64

well, I know a = -1/8 regardless of what that other stuff does :)

31. amistre64

and b=-f ...

32. amistre64

i spose we gotta undo the cosines now and have some systems of equations to play with for the remaing terms

33. amistre64

(-8f) x^2+(-8g+8a)x+(-8h+8b+2f)=0x^2+0x+0 f = 0 so b = 0

34. amistre64

which means h=0 too for that constant

35. amistre64

(-8(0)) x^2+(-8g+8a)x+(-8(0)+8(0)+2(0))=0x^2+0x+0 -8g+8a = 0; and a = -1/8 -8g-1 = 0 ; g = -1/8

36. amistre64

2(-1/8)-8c-4(-1/8)=-3 -1/4 -8c +1/2=-3 -8c =-3' 1/4 c =13/32

37. amistre64

$y''-4y=(x^2-3)sin(2x)$ $y=c_1e^{2x}+c_2e^{-2x}-(\frac{1}{8}x^2-\frac{13}{32})sin(2x)-\frac{1}{8}xcos(2x)$

38. amistre64

and i believe the wolf agrees .... yay!!

39. amistre64

if JamesJ had been here he could of answered it with one hand tied behind his back ;)

40. gopeder

Amistre64? is that you?

41. gopeder

remember me its gopeder :)

42. amistre64

thats what it says on my nametag :)

43. gopeder

Do you Remember me im the bear guy :)

44. amistre64

i vaguely recall a screenname like yours; at my age its all i can do to remember to wear socks

45. gopeder

xD we talked alot maybe 4 mounts ago

46. amistre64

yeah, 4 months is an eternity around here lol how you been?

47. gopeder

haha so true :) Not much Just starting Geometry.. its loads of fun

48. amistre64

if you ever meet euclid, punch him in the breadbasket for me. hes to blame for alot of that geometry stuff

49. gopeder

xD oh i will :P

50. gopeder

But at the end of the day its really cool because it relates to real world

51. bahrom7893

52. robtobey

Refer to the attachment.

53. amistre64

We got "derive" at the college that looks quite similar to that kind of thing.