What is the solution for x^{''}+9x=2sin(t) ?

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

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x^{''}+9x=2sin(t)

x^{^{''}} + 9 x = 2 sin (t)

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

Solution steps

x^{''} (t) + 9 x = 2 sin (t)

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} = 7

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = (1 + e^{- x})

y^{^{''}} - 4 y = \frac{e ^{2 x} - e ^{- 2 x}}{2}

(D^{2} + D + 1) y = e^{x} sin (x)

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