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Given the problem: $y’’+16y=\cos(4x)$

It’s particular $y$ is equal to zero. I know how to get the complementary $y$ but I had problem with the particular $y. $

\begin{align*} y_p&=A\sin(4x)+B\sin(4x)\\ y_p’&=4A\cos(4x)-4B\sin(4x)\\ y_p''&=-16A\sin(4x)-16B\cos(4x) \end{align*}

As we substitute the $y_p''$ and $y_p’$ to $y''+16y=\cos(4x),$ we get:

\begin{align*} -16A\sin(4x)-16B\cos(4x)+16(A\sin(4x)+B\sin(4x))&=\cos(4x)\\ -16A\sin(4x)-16B\cos(4x)+16A\sin(4x)+16B\sin(4x)&=\cos(4x) \end{align*}

Notice that $-16A\sin(4x)$ will cancel with $16A\sin(4x),$ the same with the $-16B\cos(4x),$ which will also cancel with $16B\cos(4x).$ The $A$ will equal $0$ and also the $B$ will equal $0.$

Is there anything wrong with the equation?

Thanks for those who could help.

Adrian Keister
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Bido262
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2 Answers2

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$y_1 = \sin(4x)$ and $y_2 = \cos(4x)$ are solutions to the homogeneous equation. So by definition they are expected to cancel out. You need to try for a particular $y$ in the form of $y_p = Ax\sin(4x) + Bx\cos(4x)$.

Da Mike
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The central problem here is that the RHS of your original DE contains only terms that are also in the homogeneous solution $$y_h(x)=A\cos(4x)+B\sin(4x). $$ Thus, the LHS is going to annihilate your choice of a particular solution. What to do? This is precisely the situation in which variation of parameters comes in, and allows you to find a particular solution that is NOT annihilated by the LHS. It can be shown that the correct ansatz for your particular solution is \begin{align*} y_p(x)&=Cx\cos(4x)+Ex\sin(4x)\\ y_p'(x)&=\sin (4 x) (-4 C x+E)+\cos (4 x) (C+4 E x)\\ y_p''(x)&=8 \cos (4 x) (-2 C x+E)-8 \sin (4 x) (C+2E x). \end{align*} So we plug in to get \begin{align*} &8 \cos (4 x) (-2 C x+E)-8 \sin (4 x) (C+2E x)\\ +&16\left[Cx\cos(4x)+Ex\sin(4x)\right]\\ =&\cos(4x). \end{align*} Set things equal and solve for the constants, and you've got your particular solution.

Adrian Keister
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