What is the solution for x^{''}+9x=sin(t)+sin(3t) ?

The solution for x^{''}+9x=sin(t)+sin(3t) is x=c_{1}cos(3t)+c_{2}sin(3t)+1/8 sin(t)-(tcos(3t))/6

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x^{''}+9x=sin(t)+sin(3t)

x^{^{''}} + 9 x = sin (t) + sin (3 t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t) - \frac{t cos ( 3 t )}{6}

Solution steps

x^{''} (t) + 9 x = sin (t) + sin (3 t)

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t) - \frac{t cos ( 3 t )}{6}

x = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{1}{8} sin (t) - \frac{t cos ( 3 t )}{6}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 5 e^{x}

y^{^{''}} - 10 y + 25 = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{1}{x ^{2}} e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{15 e ^{t}}{1 + t ^{2}}

x^{^{''}} + 4 x^{^{'}} + 29 = 0, x (0) = 5, x^{^{'}} (0) = 5

What is the solution for x^{''}+9x=sin(t)+sin(3t) ?
The solution for x^{''}+9x=sin(t)+sin(3t) is x=c_{1}cos(3t)+c_{2}sin(3t)+1/8 sin(t)-(tcos(3t))/6