Question

need help with diff. eq. h/w y'' +4y = sin^2(x) need help with diff. eq. h/w y'' +4y = sin^2(x) @Mathematics

Answer 1

I found the roots and have the complementary equations y=c1cos(2x)+c2sin(2x) where c1 and c2 are constants

Answer 2

So there are two steps here ... and you've done the first step.

Answer 3

then I took the derivative and double derivative of y

Answer 4

Now for the particular solution, I recommend using the method of undetermined coefficients.

Use a trial solution of yp = A sin^2 x + B sin x cos x + C cos^2 x

Answer 5

Okay that's what i want to do but i am supposed to use a different method. where I assume part of the first derivative of y is equal to 0 and then plug in y^I and y^II into the original equal so I end up with 2 equations with two unknowns. I think I am supposed to assume the constants are a function of x

Answer 6

I don't know what the method is called though

Answer 7

You're meant to turn this into a first order system? That's also a very good method.

Answer 8

i found 1 equation just by plugging in y'' +4y = sin^2(x) and cancelling some terms out. but I am confused how to find the second one

Answer 9

What you mean in that case is that not powers of y as you've written, but define two new variables:
y1 = y
y2 = y' = y1'

Then the equation is y'' + y = sin^2 x is equivalent to
y2' + y1 = sin^2 x

Hence the two equations of the first order system are

y1' = 0 + y2 + 0
y2' = -y1 + 0 + sin^2 x

Answer 10

not exactly the method that my teacher used in class was alil different

Answer 11

you still solve for a particular and complementary equation and add then in the end

Answer 12

Are you using perhaps the method of undetermined parameters?

Using your two complimentary solutions, call them y1 and y2 (nothing to do with the y1 and y2 of what I just wrote) and finding a particular solution of the form:

yp = u1(x) y1 + u2(x) y2
?

Answer 13

yes i think so that seem right

Answer 14

Ok, so yp = u1 . sin 2x + u2 cos 2x

Now differentiate twice and substitute into your equation. Once you've done that carefully, what do you get?

Answer 15

I ened up with -2(u1)'sin(2x)+2(u2)'cos(2x)=sin^2x

Answer 16

actually i might have u1 and u2 backwardds

Answer 17

-2(u2)'sin(2x)+2(u1)'cos(2x)=sin^2x

Answer 18

u1 and u2 are functions of x right?

Answer 19

yes, most definitely.

Answer 20

okay so if my first equation is right how do you find the second equation. I remember setting part of the first derivative of y equal to zero

Answer 21

but that doesn't seem to work

Answer 22

I don't agree with your equation above.

Answer 23

Where are the second derivatives, for instance?

Answer 24

the second derivative of y=u1cos2x+u2sin2x?

Answer 25

Where are the second derivatives of the u1 and u2 ... are you sure they cancel out?

Answer 26

yp = u1 . sin 2x + u2 cos 2x
yp' = u1' sin 2x + u2' . cos 2x +
2 u1 . cos 2x - 2 u2 sin 2x

yp'' = u1'' sin 2x + u2'' . cos 2x + 2 u1' . sin 2x - 2 u2' . cos 2x
-4 u1 . sin 2x - 4 u2 . cos 2x + 2 u1' . cos 2x - 2 u2' . sin 2x

Right?

Hence yp'' + 4 yp = sin^2 x implies that ....

Answer 27

stuff

Answer 28

ohhh okay i see whats wrong when i did it i followed the example he did in class where he took the first derivative so in general you have
y=c1(x)y(x)+c2(x)y2(x)
y'=c1'(x)y(x)+c2'(x)y2(x)+c1(x)y'(x)+c2(x)y2'(x) then what he did was assume c1'(x)y(x)+c2'(x)y2(x)=0 and take the derivative of what is left in y'

Answer 29

let me tell you the answer so you can work towards it. The particular solution is

yp = 1/6 . cos 2x + 1/2

Answer 30

okay that's helpful. But i am confused on the method I am supposed to use. My teacher had 2 equations
c1'(x)y(x)+c2'(x)y2(x)=0 and whatever you you get when you plug into y'' +4y = sin^2(x

Answer 31

where the second equation was in the form c1'(x)y'(x)+c2'(x)y2'(x)=f(x)

Answer 32

I'm honestly not sure where that assumption c1'(x)y(x)+c2'(x)y2(x)=0 comes from right now; but use it for now as it will simplify your life a lot and see if it works. I'll think about why that might be true.

Answer 33

Right.

Answer 34

yeah i don't understand why that assumption works but from what I understood from class this method works on differential equation where you cant guess solutions from the method of undetermined coefficents

Answer 35

Ok. I've got to run. But I'll think about this problem some more and comment when I come back.

Answer 36

okay thanks its not due untill tomorrow at 11am so i might add some things if I figure anything out