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

The solution for x^{''}+x=2cos(t) is x=c_{1}cos(t)+c_{2}sin(t)+tsin(t)

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

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

x = c_{1} cos (t) + c_{2} sin (t) + t sin (t)

Solution steps

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

Solve linear ODE:

x = c_{1} cos (t) + c_{2} sin (t) + t sin (t)

x = c_{1} cos (t) + c_{2} sin (t) + t sin (t)

y^{^{'''}} - y^{^{''}} = 3 e^{x} + sin (x)

y^{^{''}} - 14 \cdot y^{^{''}} + 45 \cdot y^{^{'}} = 32 e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 4 t e^{3} t, y (0) = - 5, y^{^{'}} (0) = - 3

y^{^{''}} + y^{^{'}} - 2 y = - 4, y (0) = 2, y^{^{'}} (0) = 3

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

What is the solution for x^{''}+x=2cos(t) ?
The solution for x^{''}+x=2cos(t) is x=c_{1}cos(t)+c_{2}sin(t)+tsin(t)