What is the solution for y^{''}+4/490 y^'+20y= 2/245 cos(6t) ?

The solution for y^{''}+4/490 y^'+20y= 2/245 cos(6t) is

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y^{''}+4/490 y^'+20y= 2/245 cos(6t)

y^{^{''}} + \frac{4}{490} y^{^{'}} + 20 y = \frac{2}{245} cos (6 t)

y = e^{- \frac{t}{245}} (c_{1} cos (\frac{1200499 t}{245}) + c_{2} sin (\frac{1200499 t}{245})) + \frac{3}{1920818} sin (6 t) - \frac{490}{960409} cos (6 t)

Solution steps

y^{''} (t) + \frac{4}{490} y^{'} (t) + 20 y = \frac{2}{245} cos (6 t)

Solve linear ODE:

y = e^{- \frac{t}{245}} (c_{1} cos (\frac{1200499 t}{245}) + c_{2} sin (\frac{1200499 t}{245})) + \frac{3}{1920818} sin (6 t) - \frac{490}{960409} cos (6 t)

y = e^{- \frac{t}{245}} (c_{1} cos (\frac{1200499 t}{245}) + c_{2} sin (\frac{1200499 t}{245})) + \frac{3}{1920818} sin (6 t) - \frac{490}{960409} cos (6 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 4 y = - 10 x e^{x}

y^{^{''}} + 4 y^{^{'}} + 5 = 50 x + 13 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 12 x

y^{^{''}} + 3 y^{^{'}} = 3 x

What is the solution for y^{''}+4/490 y^'+20y= 2/245 cos(6t) ?
The solution for y^{''}+4/490 y^'+20y= 2/245 cos(6t) is