# Second order ODE with product of two functions on RHS

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?

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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

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Many thanks for your reply. I haven't heard of 'method of annihilators' and operators acting on diff eqns yet. I have, however, come across variation of parameters. This is a set question and we were told to do it via method of undetermined coefficients. If I just plug $y_p', y_p''$ in, I get 2 eqns but with 4 unknowns so how can I solve? – CAF Nov 24 '12 at 19:45
@CAF upon plugging in $y_p$ into $y_p''+16y_p=64x\cos(4x)$ you should get an equation which involves $\cos(4x),x\cos(4x),x^2\cos(4x)$ and $\sin(4x),x\sin(4x),x^2\sin(4x)$ these are linearly independent functions so equating their coefficients gives you seemingly 6 independent equations. However, the details will only allow you to eliminate 4 of the 6 undetermined coefficients I mentioned. Don't forget the product rule as you calculate $y_p'$ and $y_p''$, I usually see students miss that and waste a bunch of time on meaningless equations... – James S. Cook Nov 24 '12 at 19:58
I did remember the product rule. What I did was sub in and collected all terms that had a cos4x term and all terms that had a sin4x term. I then set the former to 64x and the latter to 0. This seems correct - is it not? – CAF Nov 24 '12 at 20:32
For example. if you came to $$A\cos(4x)+(A+B)x\cos(4x)+(C+3D)\sin(4x)+(D-C)x\sin(4x) +E\cos(4x)= 64x\cos(4x)+\sin(4x)$$ then I can read that $$A=0, A+B=64, C+3D=1,D-C=0, E=0$$ of course these are not what you'll get, I just made them up. And, yes that approach you mention sounds correct, just one more step to make, if $Ax^2+Bx+C = x^2-3x$ then $A=1, B=-3, C=0$. – James S. Cook Nov 24 '12 at 20:34
What do you mean by the last statement? The eqns I get in the end are $-8(2Cx+D) + 2CB = 0$ and $2CA + 8B(2Cx+D) = 64x$ How do I solve this? – CAF Nov 24 '12 at 20:41

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: $$4x^2\sin(4x)+x\cos(4x)+a\sin(4x)+b\cos(4x)$$

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Sorry pressed enter by accident. We haven't covered vector spaces of homogeneous diff eqns in the course where this question is from. Are my methods valid? – CAF Nov 24 '12 at 17:43
Solutions of a homogenous differential equation can be added and multiplied by reals, therefore they form a vector space. It is well known an can be proved by utilising the properties of the wronskian that a n-dimensional vector space has at most a n-dimensional space of solutions, so the set of solutions given above is complete. The problem with your attempts is that they include no argument why there shouldn't be any other solutions. – Dominik Nov 24 '12 at 22:36
I nearly get your proposed soln, but I have: $y_p = x\cos4x + (4x^2 + \frac{1}{8}x)\sin4x$ – CAF Nov 25 '12 at 21:39
Wolframalpha says i'm right: wolframalpha.com/input/?i=y%27%27%2B16y%3D64xcos%284x%29 – Dominik Nov 25 '12 at 21:56
Pardon me. I made an error. Thanks! – CAF Nov 25 '12 at 22:02