Question

Q1 Find a particular solution to the nonhomogeneous differential equation y''+4y=cos (2x) + sin (2x).



Q2 Find the most general solution to the associated homogeneous differential equation. Use A and B in your answer to denote arbitrary constants.


Q3. Find the solution to the original nonhomogeneous differential equation satisfying the initial conditions y(0) = 4 and y'(0) = 2.

Y= ??

Answer 1

anyone able to help ??

Answer 2

\[\Large r^2+4=0 \]
\[\Large r^2=i^24 \longrightarrow r=\pm 2i\]

Answer 3

So I would agree,
\[\Large \cos(2x)+\sin(2x) \]
is a solution

Answer 4

yes i got that far but you have to let it equal the inital conditions but it doesnt work of for me in the end

Answer 5

Well usually you apply the initial conditions at the very last step, the way I go is usually solve the associated homogeneous differential equation first, then use the method of undetermined coefficients to solve it generally and then apply the initial conditions.

Answer 6

So a guess for the RHS by undetermined coefficients could be:
\[\Large A\sin(2x)+B\cos(2x)+C\sin(2x)+D\cos(2x) \]
Seems like we have redundant constants in here, so this seems more pleasant to deal with:
\[\Large A\sin(2x)+B\cos(2x) \]
basically merging constants together. But since the above is already a solution by the homogeneous differential equation we want to add powers of \(x\) to it:
\[\Large Ax\sin(2x)+Bx\cos(2x) \]

Answer 7

yes its just trying to get the right solution as i think you have to add Yp + Yh to get y

Answer 8

yes exactly, the principle of superposition.

Answer 9

Yes. Use undetermined coeffs

Answer 10

What did any of ye figure A and B to be ?

Answer 11

what did you get? I can double check with my answers.

Answer 12

I got A = 4 and B = 1 but not sure am i doing it right

Answer 13

I got A=1/4, B=-1/4

Answer 14

Yes i had that answer in previous part of the question i got part 1 and 2 . its part 3 i cant figure out

Answer 15

Just to make sure we're on the same level of the question, so you have the most general solution to the above given DE. Because without that you can't continue, from there you need to continue by adding constants.

\[\Large y=c_1\cos(2x)+c_2\sin(2x)+\frac{1}{4}x\sin(2x)-\frac{1}{4}x\cos(2x) \]

Answer 16

and now, and now only, I would suggest adding initial conditions to solve for \(c_1\) and \(c_2\). It will give you a system of linear equations.

Answer 17

Yes so then you let it equal the inital values and i just keep getting wrong answer

Answer 18

Well, above you mentioned something about A and B and that is what confused me. A and B are constants, but they are constants of the method called undetermined constants, not to confuse with the constants of the integration \(c_1, c_2\). To me it seems like:

\[\Large y(0)=c_1=4 \] is correct

Answer 19

yes i think i just forget to get y' of the end part of y thats why i keep getting wrong answer B =1

Answer 20

alright then (-:

Answer 21

well i am trying to get answer but must be going wrong somewhere

Answer 22

did anyone find the solution ?

Answer 23

@Spacelimbus already gave you the General Solution. Your remaining task is to find the last two paramenters. Go forth!!

Answer 24

I have did it all out but I can't seem to get solution right as I have to enter solution into computer and still won't work