如题想求解这个方程，但是一直plus标签被保护。二楼上代码谢谢大佬了。

qu1 = fu1[t]
qv1 = fv1[t]
qu = {qu1}
qv = {qv1}
M = MatrixForm[{{{34.45428319573334` + 10.472846264820031` qu +
0.8075222624000002` qu^2 -
0.3122061168793633` qv^2}, {1.0946841805411454` qv}, \
{7.348812836608349` + 1.0946841805411454` qu}},
{{-1.0946841805411454` qv}, {0.8075222624000002`},
0}, {{7.348812836608349` + 1.0946841805411454` qu},
0, {1.6147756243801052`}}}]
G = MatrixForm[-2 Dt[(4 t)/15 + 4/15 \[Pi]^2 Sin t,
t] {{0, 0, 0}, {0,
0, {1.0946841805411454`}}, {0, {-1.0946841805411454`}, 0}}]
K = MatrixForm[ {{0, 0, 0},
{0, {775888.0835571943` -
0.8075222624000002` (4/15 + (4 \[Pi]^2 Sin)/15 +
4/15 \[Pi]^2 t Dt[Sin, t])^2}, 0}, {0,
0, {13.681061451793486` +
0.3122061168793633` (4/15 + (4 \[Pi]^2 Sin)/15 +
4/15 \[Pi]^2 t Dt[Sin, t])^2}}}]
Q = MatrixForm[{{{-2 (4/15 + (4 \[Pi]^2 Sin)/15 +
4/15 \[Pi]^2 t Dt[Sin, t]) (5.236423132410016` Derivative[1][
fu1][t] +
0.8075222624000002` fu1[t] Derivative[1][fu1][t] -
0.3122061168793633` fv1[t] Derivative[1][fv1][
t])}}, {{5.236423132410016` (4/15 + (4 \[Pi]^2 Sin)/15 +
4/15 \[Pi]^2 t Dt[Sin, t])^2}}, {0}} + {{(2 \[Pi] Sin t)/
15}, {0}, {0}}]
q = MatrixForm[Transpose[( {
{{(4 t)/15 + 4/15 \[Pi]^2 Sin t}, {fu1[t]}, {fv1[t]}}
} )]]
dSolveAlways[
M Dt[q, {t, 2}] + G Dt[q, t] + K q = Q, {fu1[t], fv1[t]}]

那个方程应该是 == Q

更新一下，根据大佬的指点重新写了一遍。由于实在比较长现在吧nb文件传到了云盘求大佬们指点一下。
链接：[无效] https://pan.baidu.com/s/1QJ_jbvnMVcUMrQwhonqgnA
提取码：8ccl

Dsolve[(13.681061451793486 + 0.3122061168793633*
(4/15 + (4/15)*Pi^2*Cos[(2*Pi*x)/15])^2)*y[x] -
16.203329712612796*Sin[(2*Pi*x)/15] +
1.6147756243801052*Derivative[2][y][x] == 0, y[x], x]
现在化简成这个但是还是没有解

这是两次运行结果的代码截图，其中两次不同的只有y[x]前面的系数。
第一次的代码
DSolve[1.61478*Derivative[2][y][x] +
(13.6811 + 0.312206*(4/15 + (4/15)*Pi^2*Cos[2*Pi*(x/15)])^2)*
y[x] -
16.2033*Sin[(2*Pi*x)/15] == 0, y[x], x]
第二次的代码
DSolve[1.61478*Derivative[2][y][x] + (x + 1)*y[x] -
16.2033*Sin[(2*Pi*x)/15] == 0, y[x],
x]