Question

Need some help please.
Obtain the general solution:
(D^2+D)y=-cos x

Answer 1

is this differential equations?

Answer 2

Yes

Answer 3

what book are you using ... this looks familiar somehow....

Answer 4

My instructor does not require a text book.. Let me show you how far I've gotten..

Answer 5

oh this looks like one of those 1960s problems I had that last semester...they're the only books I know that uses D

Answer 6

Lol yeah she is a bit old school..

Answer 7

is this differential equations I or II?
I feel like I should say undetermined coefficients... because of that -cos x thing... so that would be Yp = Acosx +Bsinx but this could also be something else... like variation of parameters...or something

Answer 8

The full solution is:
\[y=C1+C2e^{-x}+\frac{ 1 }{ 2 }\cos x - \frac{ 1 }{ 2 }\sin x\]
I can get the first half but the 2nd half is where I need help.

Answer 9

Yes you're right on.. it is DE I, Linear, non-homogeneous equations by undetermined coeff.

Answer 10

ok so we need to find the Yp
and that -cosx ... means that's in the form of Yp =Acosx +Bsinx you need to find the first and second derivative of Yp and plug it into the equation... if you have 0 = something on the right
we need a multiplier like by x on Yp and do the whole process again.

Answer 11

so find the derivatives of the Yp and plug it into the D's D = 1st derivative D^2 = second derivative

Answer 12

So just to make sure, would the first derivative be:
\[y'p=A(-\sin x)+B(\cos x)\]
and the Second Der.:
\[y''p=A(-\cos x)+B(-\sin x)\]

Answer 13

yeah it should... I have to scan something... keep working at this problem

Answer 14

oh wait not scan something more like have to ummm do something.. so I'll be back in 15 minutes..so in the meantime plug in those derivatives in D and then we need to find all the cosx and sinx... all the sinx should have a 0 on the right since we don't have a sinx to deal with in the first place. but we do have a cosx and we need to solve to find our A and B.

Answer 15

the Y is confusing on the original problem... maybe if we have it as D^2+D = -cosx we can see it easier.

Answer 16

Ok im just not sure exactly how to plug them in but ill give it a shot..

Answer 17

O_O second derivative of Yp goes in D^2
first derivative of Yp goes in D

Answer 18

I'll come back.. in 10-15 minutes.... need to wash up

Answer 19

right, my bad, but then im not sure where to go from there..

Answer 20

After plugging in, I have:
\[[A(-\cos x)+B(-\sin x)]+[A(-\sin x)+B(\cos x)] = - \cos x\]

Answer 21

I just don't really know how to clean it up lol

Answer 22

First off how do you find D?\[D=\left[\begin{matrix}0 & -1 \\ 1& 0\end{matrix}\right]\] since when you multiply a vector with the basis {sinx, cosx} you will get it's derivative since the derivative is just a linear transformation! Try it now!

\[\left(\begin{matrix}3 \\ -2\end{matrix}\right)=3 \sin x -2 \cos x\]

So now if you multiply \[\left[\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right] \left(\begin{matrix}1 \\ 0\end{matrix}\right)\] this is the same thing as taking the first derivative of sin(x) right?

Answer 23

Wow I was never shown that way!

Answer 24

Similarly, D^2 will just be the negative identity matrix. No surprise there either?

Answer 25

So now add up D^2+D which is multiplied by your vector y. This equals the other vector, (0,-1) since that is just -cosx.

Since it's really easy to find the inverse of a 2x2 matrix, just multiply by the inverse of each to get y!\[\bar y = (D^2+D)^{-1}\left(\begin{matrix}0 \\ -1\end{matrix}\right)\]

Answer 26

I could give more help if you need it, I'm just giving the basic outline here of what's going on.

Answer 27

Interesting.. it was never explained that way... now im looking for my Yp and once I find it I plug back into the original equation right? after I do that I am confused on how to clean it up or how to find A and B..

Answer 28

Well that would be the slower method of doing it... But from the notation your teacher is using it suggest she wants you to do it the way I'm showing you.

Answer 29

Ok, can you show me how to do it by your method?

Answer 30

WHAT THE HECK HAPPENED HERE?

Answer 31

I had old problems like these.... I had to do it the slower method and it worked out so ... if it ain't broke don't fix it.

Answer 32

Lol this dude tried to help me but then left...
Did u see where I plugged them in?

Answer 33

\[[A(-\cos x)+B(-\sin x)]+[A(-\sin x)+B(\cos x)] = - \cos x \]

Answer 34

Yes.. just not sure how to clean up or solve..

Answer 35

\[[-A\cos x)+(-B\sin x)]+[(-A\sin x)+B(\cos x)] = - \cos x \]

Answer 36

\[-A \cos x + B cosx = -cosx \]
\[-Asinx -Bsinx = 0 \]

Answer 37

you split them up into two parts..
so factor out the cosx and divide by cosx
same goes for sinx

Answer 38

\[cosx(-A+B) = -cosx \]

Answer 39

\[sinx( -A-B) = 0\]

Answer 40

Ohhhhhhhhhhhhh.... nice

Answer 41

\[-A + B = -1\]
\[-A-B=0\]

Answer 42

\[-2A = -1 \]
\[A = \frac{1}{2}\]

Answer 43

\[-\frac{1}{2}-B = 0\]

Answer 44

\[-B =\frac{1}{2}\]
\[B = \frac{-1}{2}\]

Answer 45

I see!! Nice thanks.. I guess that was just a brain fart.. I needed to see a couple problems worked out, I've got quite a bit to do and I missed class that day because I was sick.

Answer 46

plugging those two values back into the ORIGINAL \[Y_p\] would result in \[\frac{1}{2} \cos x -\frac{1}{2}sinx\]

Answer 47

now combine that with the result of \[Y_h\] will become \[y=C1+C2e^{-x}+\frac{ 1 }{ 2 }\cos x - \frac{ 1 }{ 2 }\sin x\]

Answer 48

You're a lifesaver! Btw, whats your major if you dont mind me asking?

Answer 49

MAth :P hence the stupid refine the relation on a strict poset problem being unanswered. I got the solution on paper not sure if it's right but hey at least I tried :P

Answer 50

differential equations i is math for mechanical engineers I'm taking another differential equations course next semester that's for Physics majors. ... I think. I love the math, but hate the technical reading.

Answer 51

Gotcha! Thats awesome! Im actually majoring in Petroleum Eng. Well thanks very much for your help... Hope to see you on here soon cuz I know ill need the help again!