alka8q7

2021-11-23

I have the following differential equation:
$y+y=\mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)$
Maybe something can be done to $\mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)$ to make it easier to solve. Any ideas?

Supoilign1964

I don't see an elegant solution , but it can be done by brute force using the method of variation of parameters.
The homogeneous differential equation $y{}^{″}\left(t\right)+y\left(t\right)=0$ has the two fundamental solutions ${y}_{1}\left(t\right)=\mathrm{sin}\left(t\right)$ and ${y}_{2}\left(t\right)=\mathrm{cos}\left(t\right)$
We must solve
$\left(\begin{array}{c}0\\ \mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)\end{array}\right)=\left(\begin{array}{cc}{y}_{1}& {y}_{2}\\ {y}_{1}^{\prime }& {y}_{2}^{\prime }\end{array}\right)\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)=\left(\begin{array}{cc}\mathrm{sin}t& \mathrm{cos}t\\ -\mathrm{cos}t& \mathrm{sin}t\end{array}\right)\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)$
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
${y}_{0}\left(t\right)={U}_{1}{y}_{1}+{U}_{2}{y}_{2}$ and $y={y}_{0}+{c}_{1}{y}_{1}+{c}_{2}{y}_{2}$
will be the general solution.
Since the matrix $A\left(t\right)=\left(\begin{array}{cc}\mathrm{sin}t& \mathrm{cos}t\\ -\mathrm{cos}t& \mathrm{sin}t\end{array}\right)$ is orthogonal, its inverse is its transpose, so
$\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)=\left(\begin{array}{cc}\mathrm{sin}t& -\mathrm{cos}t\\ \mathrm{cos}t& \mathrm{sin}t\end{array}\right)\left(\begin{array}{c}0\\ \mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)\end{array}\right)=\left(\begin{array}{c}{\mathrm{cos}}^{2}\left(t\right)\mathrm{cos}\left(2t\right)\\ \mathrm{sin}\left(t\right)\mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)\end{array}\right)$
Note that ${v}_{2}\left(t\right)=\mathrm{sin}\left(t\right)\mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)=\frac{1}{2}\mathrm{sin}\left(2t\right)\mathrm{cos}\left(2t\right)=\frac{1}{4}\mathrm{sin}\left(4t\right),$ so we can take
${U}_{2}\left(t\right)=-\frac{1}{16}\mathrm{cos}\left(4t\right)$
On other hand, we get
${u}_{1}={\mathrm{cos}}^{2}\left(t\right)\mathrm{cos}\left(2t\right)=\mathrm{cos}\left(t\right)\frac{1}{2}\left(\mathrm{cos}\left(3t\right)+\mathrm{cos}\left(t\right)\right)$

Charles Randolph

There's a general way of doing these things. You solve the homogeneous equation $y{}^{″}+y=0$, giving ${C}_{1}\mathrm{cos}\left(t\right)\mathrm{sin}\left(t\right)+{C}_{2}\mathrm{sin}\left(t\right)$ for constants ${C}_{1}$ and ${C}_{2}$, then add to it a single solution ${y}_{p}\left(t\right)$ to the inhomogeneous equation $y{}^{″}+y=\mathrm{cos}\left(t\right)\mathrm{cos}\left(2t\right)=\frac{1}{2}\mathrm{cos}\left(3t\right)+\frac{1}{2}\mathrm{cos}\left(t\right)$. The resul will be the general solution to your differential equation
To find ${y}_{p}\left(t\right)$, you try ${y}_{p}\left(t\right)={a}_{1}\mathrm{cos}\left(3t\right)+{a}_{2}\mathrm{sin}\left(3t\right)+{b}_{1}t\mathrm{cos}\left(t\right)+{b}_{2}t\mathrm{sin}\left(t\right)$. You plug it in and solve for ${a}_{1},{a}_{2},{b}_{1},$ and ${b}_{2}$. Normally you just try combinations of $\mathrm{cos}\left(3t\right),\mathrm{sin}\left(3t\right),\mathrm{cos}\left(t\right)$ and $\mathrm{sin}\left(t\right)$, 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 and ${a}_{2}={b}_{1}=0$. Thus your general solution will be
$y\left(t\right)={C}_{1}\mathrm{cos}\left(t\right)+{C}_{2}\mathrm{sin}\left(t\right)-\frac{1}{16}\mathrm{cos}\left(3t\right)+\frac{1}{4}t\mathrm{sin}\left(t\right)$
(The $\frac{5}{16}\mathrm{cos}\left(t\right)$ term gets absorbed into the solution to the homogeneous equation).

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