What is the solution for y^{''}+y=1sin(18.8t) ?

The solution for y^{''}+y=1sin(18.8t) is y=c_{1}cos(t)+c_{2}sin(t)-0.00283…sin(18.8t)

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y^{''}+y=1sin(18.8t)

y^{^{''}} + y = 1 sin (18.8 t)

y = c_{1} cos (t) + c_{2} sin (t) - 0.00283 \dots sin (18.8 t)

Solution steps

y^{''} (t) + y = 1 sin (18.8 t)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - 0.00283 \dots sin (18.8 t)

y = c_{1} cos (t) + c_{2} sin (t) - 0.00283 \dots sin (18.8 t)

D (x) = 0.2 x^{2} + 9.2 x - 1.3

y^{^{''}} + 5 y^{^{'}} = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 x + 5, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 + e^{x}

y^{^{''}} + 2 y = - 3

What is the solution for y^{''}+y=1sin(18.8t) ?
The solution for y^{''}+y=1sin(18.8t) is y=c_{1}cos(t)+c_{2}sin(t)-0.00283…sin(18.8t)