What is the solution for x^{''}+4x=sin(5t) ?

The solution for x^{''}+4x=sin(5t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/21 sin(5t)

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x^{''}+4x=sin(5t)

x^{^{''}} + 4 x = sin (5 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{21} sin (5 t)

Solution steps

x^{''} (t) + 4 x = sin (5 t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{21} sin (5 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{21} sin (5 t)

y^{^{''}} + y = sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = tan (3 t)

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 e^{x} - 18 e^{- x}

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

y^{^{''}} - 2 y^{^{'}} + y = e^{2 t}

What is the solution for x^{''}+4x=sin(5t) ?
The solution for x^{''}+4x=sin(5t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/21 sin(5t)