1
$\begingroup$

Find the general solution of the ODE $y′′ +16y=64x\cos4x.$ If $y(0)=1, y′ (0)=0,$ what is the particular solution?

Attempt: I am just needing some help with the particular integral. I have tried it two different ways, getting to a stage where I can go no further. In the first attempt, I said $ y_p(x) = x[A\cos4x+B\sin4x](Cx+D)$ and got to the two eqns $-8A(2Cx + D) + 2CB = 0$ and the other $2CA + 8B(2Cx+D)=64x$, which is 2 eqns with 4 unknowns so can't solve to get unique constants. In my other attempt, I let $y_p(x) = x[(Ax +B)\cos4x + (Ax+B)\sin4x]$ so I would have only two constants, but in the end I get A being a function of x. Attempt 1 seems more plausible, yet I can't seem to solve the eqns. Any advice?

2 Answers 2

1

Your problem is $(D^2+16)[y] = 64x\cos(4x)$ where $D = d/dx$. Operate on both sides by $(D^2+16)^2$ to see $(D^2+16)^3[y]=0$ is the corresponding homogeneous equation from the method of annihilators. It follows that $ y = (Ax^2+Bx+C)\cos(4x)+(Ex^2+Fx+G)\sin(4x)$ is the general solution to the initial problem. You just plug this in and determine $A,B,E,F$ as particular constants and $C,G$ serve as the usual $c_1,c_2$ of the homogeneous solution.

The previous answer gave you the answer, I give you all possible other answers and how to derive said answer. That said, perhaps you are supposed to use the method of variation of parameters? This applies and will derive the needed solution from integration of certain formulas involving the fundamental solution set $\{ \cos(4x),\sin(4x) \}$.

Ultimately, this is why your initial attempt failed. The forcing term $64x\cos(4x)$ contains a homogeneous solution. It overlaps. The method of annihilators gives us the way to adjust the naive particular solution $y_p = A\cos(4x)+B\sin(4x)$ as to provide nontrivial output for the operator $D^2+16$.

Incidentally, variation of parameters here is a bit involved (this is related, but in a sense easier than your problem): Differential equation with a constant in it

  • 0
    I recombined my initial attempt at the integral into something like $y_p = (Cx^2+Dx)\cos4x + (Ex^2 +Fx)\sin4x$ which is easier to solve (I get an answer close to that suggested by Dominik. Thanks for all your help!2012-11-25
1

One can guess $4x^2\sin(4x)+x\cos(4x)$ as a correct solution. Since the two dimesional vector space of the homogenous problem is spanned by $\sin(4x)$ and $\cos(4x)$, the general solution is given by: $4x^2\sin(4x)+x\cos(4x)+a\sin(4x)+b\cos(4x)$

  • 0
    Pardon me. I made an error. Thanks!2012-11-25