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I have the following differential equation: $$y''+y=\cos(t)\cos(2t)$$

Maybe something can be done to $\cos(t)\cos(2t)$ to make it easier to solve. Any ideas?

Thanks in advance.

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Would a sum of cosines be easier than a product of cosines? –  Arturo Magidin Feb 28 '11 at 20:12
Sure! What are you thinking of? –  aesede Feb 28 '11 at 20:23
$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$; $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$. So $\cos(a)\cos(b) = \frac{1}{2}(\cos(a+b) + \cos(a-b))$. In particular, $\cos(t)\cos(2t) = \frac{1}{2}(\cos(3t) + \cos(t))$. –  Arturo Magidin Feb 28 '11 at 20:30
Do you now manage to solve it? –  Fabian Feb 28 '11 at 21:56
Thanks, Arturo! That was really helpful. –  aesede Mar 1 '11 at 13:54

2 Answers 2

I don't see an elegant solution (and I fail to see how Arturo's comment helps immediately), but it can be done by brute force using the method of variation of parameters.

The homogeneous differential equation $y''(t) + y(t) = 0$ has the two fundamental solutions $y_{1}(t) = \sin{(t)}$ and $y_{2}(t) = \cos(t)$.

We must solve \[ \begin{pmatrix} 0 \\ \cos(t)\cos(2t) \end{pmatrix} = \begin{pmatrix} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{pmatrix} \begin{pmatrix} u_{1} \\ u_{2} \end{pmatrix} = \begin{pmatrix} \sin{t} & \cos{t} \\ -\cos{t} & \sin{t} \end{pmatrix} \begin{pmatrix} u_{1} \\ u_{2} \end{pmatrix} \] for $u_{1}$ and $u_{2}$ and find $U_{1}$ and $U_{2}$ such that $U_{1}^{\prime} = u_{1}$ and $U_{2}^{\prime} = u_{2}$.

A particular solution will then be given by \[y_{0}(t) = U_{1} y_{1} + U_{2}y_{2}\qquad\text{and} \qquad y=y_{0} + c_{1} y_{1} + c_{2}y_{2}\] will be the general solution.

Since the matrix $A(t) = \begin{pmatrix} \sin{t} & \cos{t} \\\ -\cos{t} & \sin{t} \end{pmatrix}$ is orthogonal, its inverse is its transpose, so \[ \begin{pmatrix} u_{1} \\ u_{2} \end{pmatrix}= \begin{pmatrix} \sin{t} & -\cos{t} \\ \cos{t} & \sin{t} \end{pmatrix} \begin{pmatrix} 0 \\ \cos(t)\cos(2t) \end{pmatrix} = \begin{pmatrix} \cos^2(t)\cos(2t) \\ \sin(t)\cos(t)\cos(2t) \end{pmatrix} \] Note that $u_{2}(t) = \sin(t)\cos(t)\cos(2t) = \frac{1}{2}\sin{(2t)}\cos{(2t)} = \frac{1}{4}\sin{(4t)}$, so we can take

\[U_{2}(t) = -\frac{1}{16}\cos{(4t)}.\]

On the other hand, using Arturo's comment three times, we get \begin{align*} u_{1}(t) & = \cos^2(t)\cos(2t) = \cos(t)\frac{1}{2}(\cos(3t)+\cos(t)) \\ & = \frac{1}{2} \cos{(t)}\cos(3t) + \frac{1}{2} \cos(t)\cos(t) = \frac{1}{4} (\cos{(4t)} + \cos{(2t)}) + \frac{1}{4}(\cos{(2t)} + 1) \\ & = \frac{1}{4}\cos{(4t)} + \frac{1}{2} \cos{(2t)} + \frac{1}{4}. \end{align*} Integrating this gives

\[U_{1}(t) = \frac{1}{16} \sin(4t) + \frac{1}{4}\sin{(2t)} + \frac{1}{4}t.\]

Now we can simplify the particular solution

\begin{align*} y_{0} = U_{1} y_{1} + U_{2}y_{2} = \cdots = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t) \end{align*}

using some further trig identities (I'm really growing tired of this, sorry).

You should at least check yourself that $y_{0}^{\prime\prime} + y_{0} = \frac{1}{2}(\cos(t) + \cos(3t)) = \cos(t)\cos{(2t)}$ by Arturo's comment again.

Therefore the general solution is \[ y(t) = y_{0} + c_{1} y_{1} + c_{2} y_{2} = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t) + c_{1}\sin{(t)} + c_{2} \cos(t) \] for some real constants $c_{1}$ and $c_{2}$.

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Your approach is the variation of parameters method ($y_p=\nu_1y_1+\nu_2y_2$, solve for $\nu_1'$ and $\nu_2'$ and integrate), which works for any inhomogeneous equation. Arturo's re-writing of $\cos(t)\cos(2t)$ as $\frac{1}{2}(\cos(3t) + \cos(t))$ allows one to use the much simpler method of undetermined coefficients. –  Eric Nitardy Mar 1 '11 at 0:12
@Eric, ah, ok. Thanks! That could have saved some work... –  t.b. Mar 1 '11 at 0:52
@Eric, if you want to fill in the details in a separate answer, feel free to do so. I'm not going to do it. –  t.b. Mar 1 '11 at 1:07

There's a general way of doing these things. You solve the homogeneous equation $y'' + y = 0$, giving $C_1\cos(t) + C_2\sin(t)$ for constants $C_1$ and $C_2$, then add to it a single solution $y_p(t)$ to the inhomogeneous equation $y'' + y = \cos(t)\cos(2t) = {1 \over 2}\cos(3t) + {1 \over 2}\cos(t)$. The result will be the general solution to your differential equation.

To find $y_p(t)$, you try $y_p(t) = a_1\cos(3t) + a_2\sin(3t) + b_1t\cos(t) + b_2t\sin(t)$. You plug it in and solve for $a_1,a_2,b_1,$ and $b_2$. Normally you just try combinations of $\cos(3t), \sin(3t), \cos(t),$ and $\sin(t)$, but since the latter two solve the homogeneous equation you have to stick a $t$ in front. I will trust Theo Buehler is right and that $a_1 = -{1 \over 16}$, $b_2 = {1 \over 4}$ and $a_2 = b_1 = 0$. Thus your general solution will be $$y(t) = C_1\cos(t) + C_2\sin(t) -{1 \over 16}\cos(3t) + {1 \over 4}t\sin(t)$$ (The ${5 \over 16}\cos(t)$ term gets absorbed into the solution to the homogeneous equation).

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Thanks for giving this much easier solution and the general hint, I was about to add it now and I am glad not to have to do it. –  t.b. Mar 1 '11 at 9:45
Thank you, Zarrax! –  aesede Mar 3 '11 at 13:06

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