Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
DIFFERENTIAL EQUATION:
help please, anybody.
(D^2 +4)y =4sin^2 (x)
 one year ago
 one year ago
DIFFERENTIAL EQUATION: help please, anybody. (D^2 +4)y =4sin^2 (x)
 one year ago
 one year ago

This Question is Closed

shubhamsrgBest ResponseYou've already chosen the best response.0
guess that substn should work..
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
aaa. i think i got it. wait, i'll try.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
ohh wait,,i didnt see a "y" also in there. !!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
@shubhamsrg I think it's just written in 'operator' notation... \[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\] correct @lilMissMindset ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
there's a y also with D^2 .. i had got that,,i didnt see a y being multiplied to whole that time..sorry..
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\] rather.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
yes he's correct I believe y =a*cos^2 x
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
did you get it to work?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
that's the particular soln. anyway...
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
since sinx is twice? applying case 4 of undetermined coefficients
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
where it has some repeated imaginary root
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
look d^2 (a*cos^2(x)) /dx^2 +4a cos^2(x) = 4sin^2(x) second derivative of a*cos^2(x) is 2a*cos^2(x)  2a*sin^2(x)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
this is for the particular solution...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
finding the homogeneous solution shouldn't be a problem...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
\[\lambda ^2 +4 =0\]
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
i'll work on it. please guide me still, in case i get stuck.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
hold on... can you tell me what part is giving you problems?
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
how about if the given is cos^x, then Yp=asin^2 x?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
if the forcing term is cos^2 x you mean?
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
i forgot 2, lol.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
yes. it would. just checked it.
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
darn, i can't do it. show me a brief solution please. so i can learn how.
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
just for the Yp part. swear.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
well, I did it already (a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
2a*cos^2(x)  2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
hmm, I see your point ...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
let me try it out and see...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
seems to work... I got a= 2 b=2
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
so 2cos^2 x 2sin^2 x
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
answer here in the book is 1/2(1xsin2x) for the Yp part.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
I'll assume that's (1/2)*( 1  x*sin^2 x )
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
you should check the original, I think you might have written the problem wrong...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
you left out and x I think...
 one year ago

lilMissMindsetBest ResponseYou've already chosen the best response.0
i don't think so.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
(1/2)*( sin^2 x  x*sin 2x ) ??
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
I don't see how (1/2)(1xsin2x) can work... maybe I'm missing something.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
something's off in your original equation or that soln.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.