Question

Solve the second order DE
\[y^{\prime\prime}+4y=\cos2x\]

Answer 1

\[y_c^{\prime\prime}+4y_c=0\]\[m^2+4=0\]\[m=\pm2i\]
\[y_c=A\cos2x+B\sin {2x}\]

Answer 2

\[y_p^{\prime\prime}+4y_p=\cos2x\]
\[y_p=C\cos2x+D\sin2x\]\[y_p^\prime=-2C\sin2x+2D\cos2x\]\[y_p^{\prime\prime}=-4C\cos2x-4D\sin2x\]

Answer 3

or was i ment to multiply by x for some reason?

Answer 4

The particular solution your looking for is yp= Ax sin(2x) because you are driving the system at its resonant frequency(cos (2x) is a solution of the homogeneous equation.)

Answer 5

can you explain further

Answer 6

Hi Everyone!

How are you all?

Answer 7

i was just following a table

Answer 8

ohhh okay

Answer 9

hi there

Answer 10

do you know of a better table?

Answer 11

@UnkleRhaukus

I want to say you that you must try again by your own self.

When we try to solve any differential equation then complementary solution is much easy to find without any problem just like algebra.

But for particular solution there are several methods to find this part. I recommend you to use "Variation of Parameters", the most powerful method to find particular solution.

Answer 12

Use Variation of Parameters. If you get command on this method then you will never get help for others.

Answer 13

yeaa UNK you were right since the particular solution is already in the form of the of the complimentary you need to multiply by x

Answer 14

\[y _{p}= x(Ccos(2x)+Dsin(2x))\]

Answer 15

since Acos(2x) was already in the compliment

Answer 16

Then you distribute the x. Take derivatives like you did before. There will be a few product rules. Plug the derivatives back in to the diff. You should get C =0 D = 1/4

Answer 17

Here's a quick and dirty derivation. Start with the problem

\[y \prime \prime + 4 y = \cos (\Omega x)\]

a particular solution(remember that there is an infinite set of particular solutions
,not just one) is then

\[y _{p}= (\cos \Omega x -\cos 2x) / (4-\Omega^2)\]

If \[\Omega = 2\] , this is indeterminate (0/0) so apply L'hopital's rule
as \[ \Omega -> 2\] to get

\[y _{p}= x \sin(2x)/4\]

Answer 18

woah that method looks cool... In my diffs class we never learned that method.

Answer 19

\[y_p=x(C\cos2x+D\sin2x)\]
\[y_p^\prime=x(-2C\sin2x+2D\cos2x)+C\cos2x+D\sin2x\]
\[y_p^{\prime\prime}=(-2C\sin2x+2D\cos2x)+x(-4Cx\cos2x-4Dx\sin2x)\]\[\qquad\qquad-2C\sin2x+2D\cos 2x\]

Answer 20

looks good...

Answer 21

\[y_p^{\prime\prime}=4(-C\sin2x+D\cos2x)+4x(-Cx\cos2x-Dx\sin2x)\]

Answer 22

\[\small(-4C\sin2x+4D\cos2x)+x(-4Cx\cos2x-4Dx\sin2x)+4x(C\cos2x+D\sin2x)=\cos2x\]

Answer 23

oh good, i might be ok from here, ill let you know if i get stuck again,

Answer 24

yea these sometimes get hairy, haha gotta love diffs

Answer 25

\[(-4C-4Dx^2+4Dx)\sin2x+(4D-4Cx^2+4Cx)\cos2x=\cos2x\]
\[(C+Dx^2-Dx)=0;\qquad (D-Cx^2+Cx)=1\]
\[C+Dx(x+1)=0;\qquad D-Cx(x+1)=1\]
\[D=1+Cx(x+1);\qquad C+\left(1+Cx\left(x+1\right)\right)x\left(x+1\right)=0\]

Answer 26

how am ment to solve for these constants?

Answer 27

Usually a bunch of the terms cancel

Answer 28

\[(-4C\sin2x+4D\cos2x)+x(-4Cx\cos2x-4Dx\sin2x)+4x(C\cos2x+D\sin2x)=\cos2x\]

Answer 29

\[\small(-4C\sin2x+4D\cos2x)+x(-4Cx\cos2x-4Dx\sin2x)+4x(C\cos2x+D\sin2x)=\cos2x\]

Answer 30

i dunno where to cancel

Answer 31

there's probably a little error let me go back through real quick

Answer 32

You were good righht up to the check mark. There shouldn't be any terms with an x^2, just solve the equation you had at that point. (the coeffs of sin and cos need to vanish separately.)

Answer 33

\[\small(-4C\sin2x+4D\cos2x)+x(-4Cx\cos2x-4Dx\sin2x)+4x(C\cos2x+D\sin2x)=\cos2x\]
silly me , this step is easy
\[(-4C\sin2x+4D\cos2x)=\cos2x\]

Answer 34

It's always the little things that mess us up....

Answer 35

\[y'' _{p} = -4Cxcos(2x) +4Dcos(2x) -4Dxsin(2x)-4Csin(2x)\]

Answer 36

=)

Answer 37

\[-4C=0;\qquad 4D=1\]\[C=0;\qquad D=\frac 14\]\[y_p=\frac x4\cos2x\]

Answer 38

not quite.....

Answer 39

ah im out of of phase for some reason

Answer 40

go pack tp yp, what does D multiply?

Answer 41

oh haha

Answer 42

\[y_p=x(C\cos2x+D\sin2x)\]
\[y_p=\frac x4\sin2x\]

Answer 43

\[y=y_c+y_p\]

\[{y (x)= A\cos2x+B\sin2x}+\frac{1}4x\sin2x\quad \Large\color\red\checkmark\]

Answer 44

Thank you all ever so much ,

Answer 45

YES!