Is my book wrong?(constant coefficients)

[schattenjaeger]

y''+16y=24cos(4x)

it was mentioned in class that our book's answer was wrong on some problem, but I forgot which one and now that's biting me in the butt(I really should've written it down)

For the particular solution I got y=3x*sin(4x), which if I plug it in works. The book's solution: (this is the general solution, so the first part is the complementary solution which is right)

y=c1cos(4x)+c2sin(4x)+1/17*x(24*cos(4x)+96*sin(4x))

 

[schattenjaeger]

Also, on the same subject if you have a DE y''-y'-2y=40sin(x)^2, what's the trail solution for it?

I know the multiple of forms on the RHS gives you product of their trial solutions, but how does this work?

since the trial solution to sin(x) would be asin(x)+bcos(x), would you throw in another csin(x)+dcos(x) for the other sin, then multiply them, and (after combining constants, which I think you can do at this step)get Asin^2(x)+Bsinxcosx+Ccos^2(x)?

I ask 'cuz I worked it out as such and got the wrong answer, went over it, found a mistake, and still got the wrong answer, and I want to know if I went wrong before then

 

[HallsofIvy]

Your solution is correct, not the textbooks. Since there is no odd derivative in the equation, you don't need both sine and cosine.

As for the second equation, do you have any reason to think that "undetermined coefficients" with any trial solution will work? Sin² x is NOT one of the functions that can satisfy a linear equation with constant coefficients.

Try using the trig identity sin²(x)= (1/2)(1- cos(2x)) to simplify the right hand side, THEN look for a trial solution.

 

[schattenjaeger]

OH

Yet another example of me being almost good at what I'm doing:-D