What is the solution for y^{''}+9y=-sin(3t),y(0)=0,y^'(0)= 1/6 ?

The solution for y^{''}+9y=-sin(3t),y(0)=0,y^'(0)= 1/6 is y=(tcos(3t))/6

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y^{''}+9y=-sin(3t),y(0)=0,y^'(0)= 1/6

y^{^{''}} + 9 y = - sin (3 t), y (0) = 0, y^{^{'}} (0) = \frac{1}{6}

y = \frac{t cos ( 3 t )}{6}

Solution steps

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

Solve linear ODE:

y = \frac{t cos ( 3 t )}{6}

y = \frac{t cos ( 3 t )}{6}

\int (x^{2} + 2 x) cos (5 x) d x

n = 1 \sum \infty 2 (0.8)^{n}

x \to 3 + lim (\frac{2}{( x - 3 ) ^{3}})

1 + \frac{1}{x ^{2}}

\int (2 x - 3 x^{2}) d x

What is the solution for y^{''}+9y=-sin(3t),y(0)=0,y^'(0)= 1/6 ?
The solution for y^{''}+9y=-sin(3t),y(0)=0,y^'(0)= 1/6 is y=(tcos(3t))/6