Question

Differential help

(D^2-1)y = 10sin(3x)

Answer 1

im trying to find my Yp

Answer 2

I struggle a bit with this notation, not really used to Euler:
\[\Large y''-y=10\sin(3x) \] ?

Answer 3

ive set up my equation to be -90Acos(3x)-90Bsin(3x)=10sin(3x)

I just need to solve this for A and B

Answer 4

im not sure what to set my equilibrium equations equal to

Answer 5

What did you take as a guess?

Answer 6

\[\Large A\sin(3x)+B\cos(3x) \]?

Answer 7

Because that would change your differentiation a bit, you should end up at:\[\Large -10A\sin(3x)-10B\cos(3x)=10\sin(3x) \]

Answer 8

but you then have to take the derivative of this twice and when you plug that into D^2-1 that should end you up where i was at

Answer 9

Well, as I requested in my opening post; I really don't see the Euler Notation too often, but if I understand this problem correctly then it's asking for:
\[\Large y''-y=10\sin(3x) \]
is this correct?

Answer 10

Or did I misinterpret the Euler Notation?
I think I remember that
\[\Large D^2f=f'' \]

Answer 11

what did you get for \(y_h\) ?

Answer 12

i dont even know what yh is

Answer 13

yc = ae^x+be^(-x)

Answer 14

yeah y_h is the same as y_c

Answer 15

ok that makes sense so how do i solve for my A and B values?

Answer 16

\[\Large y''-y=0 \\ \Large \lambda^2-1=0 \]
So I agree But your guess above still seems wrong to me.

\[\Large y_p=A\sin(3x)+B\cos(3x) \]
Differetiate that twice, I don't see how your 90 shows up.

Answer 17

im not sure how you came up with this but go ahead and continue

Answer 18

\[\Large y_p'=3A\cos(3x)-3B\sin(3x) \\
\Large y_p''=-9A\sin(3x)-9B\cos(3x) \]
Such that:
\[\Large y_p''-y_p=-10A\sin(3x)-10B\cos(3x)=10\sin(3x) \]

Answer 19

B=0, A=-1

Answer 20

wait can you go over the solving part

Answer 21

can you tell me where you've problems? At the guess? At the differentiation? Or at the last step where you try to figure out the constants.

Answer 22

the last step. solving for the A and B variable.

Answer 23

Well basically all you do is set things equal:
\[\Large -10A\sin(3x)-10B\cos(3x)=10\sin(3x) \]
Remember that the equal sign means the LHS (Left hand side) must be equal to the RHS. So on the right hand side we don't have a cos(3x) term, how can we make it disappear from the left hand side? By settign B=0, Now we're left with:
\[\Large -10A\sin(3x)=10\sin(3x) \]
Solving for A gives A=-1

Answer 24

so can just assume b to be 0? you dont have to prove this?

Answer 25

Have you done partial fraction decomposition before?

Answer 26

There you do basically the exact same thing, the idea behind this kind of differential equations it that you first solve the homogenous DE, which is usually easy to solve. Then you make a guess for the RHS.

This step is necessary because a differential equation is defined through the differentials, so in anyway, the function cannot just completely disappear. Your guess will always include unknown constants, and you're trying to solve for these. You do this by plugging in your guess into the original DE and then make your coefficients match the RHS of the equation.

In general case, this is proof enough, it's a basic concept of mathematics to match coefficients.

For further examples see:
Laws of Vieta
Integrating Factors