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

The solution for y^{''}+y=tcos(2t) is y=c_{1}cos(t)+c_{2}sin(t)-(tcos(2t))/3+4/9 sin(2t)

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

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

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t cos ( 2 t )}{3} + \frac{4}{9} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t cos ( 2 t )}{3} + \frac{4}{9} sin (2 t)

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t cos ( 2 t )}{3} + \frac{4}{9} sin (2 t)

y^{^{''}} + 9 y = sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 5 y = x sin (x)

y^{^{''}} + 25 y = 4 cos (5 x) - 7 e^{5 x}

5 y^{^{''}} + y^{^{'}} = - 4 x, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 2 y + y = e^{- t} ln (t)

What is the solution for y^{''}+y=tcos(2t) ?
The solution for y^{''}+y=tcos(2t) is y=c_{1}cos(t)+c_{2}sin(t)-(tcos(2t))/3+4/9 sin(2t)