What is the solution for y^{''}+y=3sin(2t)+2tcos(2t) ?

The solution for y^{''}+y=3sin(2t)+2tcos(2t) is y=c_{1}cos(t)+c_{2}sin(t)-(2tcos(2t))/3-1/9 sin(2t)

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y^{''}+y=3sin(2t)+2tcos(2t)

y^{^{''}} + y = 3 sin (2 t) + 2 t cos (2 t)

y = c_{1} cos (t) + c_{2} sin (t) - \frac{2 t cos ( 2 t )}{3} - \frac{1}{9} sin (2 t)

Solution steps

y^{''} (t) + y = 3 sin (2 t) + 2 t cos (2 t)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - \frac{2 t cos ( 2 t )}{3} - \frac{1}{9} sin (2 t)

y = c_{1} cos (t) + c_{2} sin (t) - \frac{2 t cos ( 2 t )}{3} - \frac{1}{9} sin (2 t)

y^{^{''}} - 16 y = sinh (4 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} cos (x) + x e^{- x}

x^{^{''}} + 4 x = sin^{2} (2 t)

y^{^{'''}} + 4 y^{^{'}} = e^{x} cos (2 x)

y^{^{''}} + 3 y^{^{'}} + 6 y = 4 e^{- 3 t} sin (t)

What is the solution for y^{''}+y=3sin(2t)+2tcos(2t) ?
The solution for y^{''}+y=3sin(2t)+2tcos(2t) is y=c_{1}cos(t)+c_{2}sin(t)-(2tcos(2t))/3-1/9 sin(2t)