I have the following differential equation: y''+y=\cos(t)\cos(2t) Maybe something can be done

alka8q7 2021-11-23 Answered
I have the following differential equation:
\(\displaystyle{y}{''}+{y}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\)
Maybe something can be done to \(\displaystyle{\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\) to make it easier to solve. Any ideas?

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Expert Answer

Supoilign1964
Answered 2021-11-24 Author has 95 answers

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 \(\displaystyle{y}{''}{\left({t}\right)}+{y}{\left({t}\right)}={0}\) has the two fundamental solutions \(\displaystyle{y}_{{1}}{\left({t}\right)}={\sin{{\left({t}\right)}}}\) and \(\displaystyle{y}_{{2}}{\left({t}\right)}={\cos{{\left({t}\right)}}}\)
We must solve
\[\left(\begin{array}{c}0\\ \cos(t)\cos(2t)\end{array}\right)=\left(\begin{array}{c}y_1&y_2\\ y_1'&y_2'\end{array}\right)\left(\begin{array}{c}u_1\\ u_2\end{array}\right)=\left(\begin{array}{c}\sin t&\cos t\\ -\cos t&\sin t\end{array}\right)\left(\begin{array}{c}u_1\\ u_2\end{array}\right)\]
for \(\displaystyle{u}_{{1}}\) and \(\displaystyle{u}_{{2}}\) and find \(\displaystyle{U}_{{1}}\) and \(\displaystyle{U}_{{2}}\) such that \(\displaystyle{U}_{{1}}'={u}_{{1}}\) and \(\displaystyle{U}_{{2}}'={u}_{{2}}\)
A particular solution will then be given by
\(\displaystyle{y}_{{0}}{\left({t}\right)}={U}_{{1}}{y}_{{1}}+{U}_{{2}}{y}_{{2}}\) and \(\displaystyle{y}={y}_{{0}}+{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}\)
will be the general solution.
Since the matrix \(A(t)=\left(\begin{array}{c}\sin t&\cos t\\-\cos t&\sin t\end{array}\right)\) is orthogonal, its inverse is its transpose, so
\[\left(\begin{array}{c}u_1\\u_2\end{array}\right)=\left(\begin{array}{c}\sin t&-\cos t\\ \cos t&\sin t\end{array}\right)\left(\begin{array}{c}0\\ \cos(t)\cos(2t)\end{array}\right)=\left(\begin{array}{c}\cos^2(t)\cos(2t)\\\sin(t)\cos(t)\cos(2t)\end{array}\right)\]
Note that \(\displaystyle{v}_{{2}}{\left({t}\right)}={\sin{{\left({t}\right)}}}{\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}={\frac{{{1}}}{{{2}}}}{\sin{{\left({2}{t}\right)}}}{\cos{{\left({2}{t}\right)}}}={\frac{{{1}}}{{{4}}}}{\sin{{\left({4}{t}\right)}}},\) so we can take
\(\displaystyle{U}_{{2}}{\left({t}\right)}=-{\frac{{{1}}}{{{16}}}}{\cos{{\left({4}{t}\right)}}}\)
On other hand, we get
\(\displaystyle{u}_{{1}}={{\cos}^{{2}}{\left({t}\right)}}{\cos{{\left({2}{t}\right)}}}={\cos{{\left({t}\right)}}}{\frac{{{1}}}{{{2}}}}{\left({\cos{{\left({3}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\cos{{\left({t}\right)}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{1}}}{{{2}}}}{\cos{{\left({t}\right)}}}{\cos{{\left({t}\right)}}}={\frac{{{1}}}{{{4}}}}{\left({\cos{{\left({4}{t}\right)}}}+{\cos{{\left({2}{t}\right)}}}\right)}+{\frac{{{1}}}{{{4}}}}{\left({\cos{{\left({2}{t}\right)}}}+{1}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\cos{{\left({4}{t}\right)}}}+{\frac{{{1}}}{{{2}}}}{\cos{{\left({2}{t}\right)}}}+{\frac{{{1}}}{{{4}}}}\)
Integrating this gives
\(\displaystyle{U}_{{1}}{\left({t}\right)}={\frac{{{1}}}{{{16}}}}{\sin{{\left({4}{t}\right)}}}+{\frac{{{1}}}{{{4}}}}{\sin{{\left({2}{t}\right)}}}+{\frac{{{1}}}{{{4}}}}{t}\)
Now we can simplify the particular solution
\(\displaystyle{y}_{{0}}={U}_{{1}}{y}_{{1}}+{U}_{{2}}{y}_{{2}}=\ldots={\frac{{{1}}}{{{4}}}}{t}{\sin{{\left({t}\right)}}}-{\frac{{{1}}}{{{16}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{5}}}{{{16}}}}{\cos{{\left({t}\right)}}}\)
using some further trig identities
You should at least check yourself that \(\displaystyle{y}{''}_{{0}}+{y}_{{0}}={\frac{{{1}}}{{{2}}}}{\left({\cos{{\left({t}\right)}}}+{\cos{{\left({3}{t}\right)}}}\right)}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\)
Therefore the general solution is
\(\displaystyle{y}{\left({t}\right)}={y}_{{0}}+{c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}={\frac{{{1}}}{{{4}}}}{t}{\sin{{\left({t}\right)}}}-{\frac{{{1}}}{{{16}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{5}}}{{{16}}}}{\cos{{\left({t}\right)}}}+{c}_{{1}}{\sin{{\left({t}\right)}}}+{c}_{{2}}{\cos{{\left({t}\right)}}}\)
for some real constants \(\displaystyle{c}_{{1}}\) and \(\displaystyle{c}_{{2}}\).

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Charles Randolph
Answered 2021-11-25 Author has 1623 answers
There's a general way of doing these things. You solve the homogeneous equation \(\displaystyle{y}{''}+{y}={0}\), giving \(\displaystyle{C}_{{1}}{\cos{{\left({t}\right)}}}{\sin{{\left({t}\right)}}}+{C}_{{2}}{\sin{{\left({t}\right)}}}\) for constants \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\), then add to it a single solution \(\displaystyle{y}_{{p}}{\left({t}\right)}\) to the inhomogeneous equation \(\displaystyle{y}{''}+{y}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}={\frac{{{1}}}{{{2}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{1}}}{{{2}}}}{\cos{{\left({t}\right)}}}\). The resul will be the general solution to your differential equation
To find \(\displaystyle{y}_{{p}}{\left({t}\right)}\), you try \(\displaystyle{y}_{{p}}{\left({t}\right)}={a}_{{1}}{\cos{{\left({3}{t}\right)}}}+{a}_{{2}}{\sin{{\left({3}{t}\right)}}}+{b}_{{1}}{t}{\cos{{\left({t}\right)}}}+{b}_{{2}}{t}{\sin{{\left({t}\right)}}}\). You plug it in and solve for \(\displaystyle{a}_{{1}},{a}_{{2}},{b}_{{1}},\) and \(\displaystyle{b}_{{2}}\). Normally you just try combinations of \(\displaystyle{\cos{{\left({3}{t}\right)}}},{\sin{{\left({3}{t}\right)}}},{\cos{{\left({t}\right)}}}\) and \(\displaystyle{\sin{{\left({t}\right)}}}\), 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 \(\displaystyle{a}_{{1}}=-{\frac{{{1}}}{{{16}}}},\ {b}={\frac{{{1}}}{{{4}}}}\) and \(\displaystyle{a}_{{2}}={b}_{{1}}={0}\). Thus your general solution will be
\(\displaystyle{y}{\left({t}\right)}={C}_{{1}}{\cos{{\left({t}\right)}}}+{C}_{{2}}{\sin{{\left({t}\right)}}}-{\frac{{{1}}}{{{16}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{1}}}{{{4}}}}{t}{\sin{{\left({t}\right)}}}\)
(The \(\displaystyle{\frac{{{5}}}{{{16}}}}{\cos{{\left({t}\right)}}}\) term gets absorbed into the solution to the homogeneous equation).
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I need some hints for solving \(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)
I noticed that the left hand side is close to \(\displaystyle{\left({y}{y}'\right)}'\):
\(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{y}{y}{''}+{\left({y}'\right)}^{{2}}-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{\left({y}{y}'\right)}'-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)
But I don't know how to continue expressing the terms as derivatives of some functions.

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