Question

DIFFERENTIAL EQUATION:
help please, anybody.

(D^2 +4)y =4sin^2 (x)

Answer 1

y= a cos2x ?

Answer 2

guess that substn should work..

Answer 3

aaa. i think i got it. wait, i'll try.

Answer 4

ohh wait,,i didnt see a "y" also in there. !!

Answer 5

@shubhamsrg
I think it's just written in 'operator' notation...
\[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\]

correct @lilMissMindset ?

Answer 6

heh.

Answer 7

there's a y also with D^2 ..
i had got that,,i didnt see a y being multiplied to whole that time..sorry..

Answer 8

\[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\]

rather.

Answer 9

yes, tat's it

Answer 10

that*

Answer 11

yes he's correct I believe y =a*cos^2 x

Answer 12

did you get it to work?

Answer 13

that's the particular soln. anyway...

Answer 14

isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx

Answer 15

why?

Answer 16

since sinx is twice? applying case 4 of undetermined coefficients

Answer 17

where it has some repeated imaginary root

Answer 18

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)

Answer 19

this is for the particular solution...

Answer 20

finding the homogeneous solution shouldn't be a problem...

Answer 21

\[\lambda ^2 +4 =0\]

Answer 22

lambda = 2i

Answer 23

ok?

Answer 24

i'll work on it. please guide me still, in case i get stuck.

Answer 25

hold on... can you tell me what part is giving you problems?

Answer 26

getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.

Answer 27

ok:)

Answer 28

how about if the given is cos^x, then Yp=asin^2 x?

Answer 29

if the forcing term is cos^2 x you mean?

Answer 30

yes.

Answer 31

i forgot 2, lol.

Answer 32

yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...

Answer 33

yes. it would. just checked it.

Answer 34

thanks!

Answer 35

sure:)

Answer 36

darn, i can't do it. show me a brief solution please. so i can learn how.

Answer 37

just for the Yp part. swear.

Answer 38

well, I did it already

(a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)

Answer 39

2a*cos^2(x) - 2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)

Answer 40

hmm, I see your point ...

Answer 41

I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)

Answer 42

let me try it out and see...

Answer 43

sure. :)

Answer 44

seems to work... I got a= 2 b=-2

Answer 45

so 2cos^2 x -2sin^2 x

Answer 46

answer here in the book is

1/2(1-xsin2x) for the Yp part.

Answer 47

I'll assume that's (1/2)*( 1 - x*sin^2 x )

Answer 48

yes.

Answer 49

you should check the original, I think you might have written the problem wrong...

Answer 50

you left out and x I think...

Answer 51

i don't think so.

Answer 52

(1/2)*( sin^2 x - x*sin 2x ) ??

Answer 53

I don't see how
(1/2)(1-xsin2x)

can work... maybe I'm missing something.

Answer 54

something's off in your original equation or that soln.