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

The solution for y^{''}+9y=cos(3t),y(pi/6)=1,y(0)=1 is y=cos(3t)+(1-pi/(36))sin(3t)+(tsin(3t))/6

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

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

y = cos (3 t) + (1 - \frac{π}{36}) sin (3 t) + \frac{t sin ( 3 t )}{6}

Solution steps

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

Solve linear ODE:

y = cos (3 t) + (1 - \frac{π}{36}) sin (3 t) + \frac{t sin ( 3 t )}{6}

y = cos (3 t) + (1 - \frac{π}{36}) sin (3 t) + \frac{t sin ( 3 t )}{6}

1.23 \cdot 2 0^{2} π 0.0 2^{2} - 0.017 y^{^{'}} = 0.01 y^{^{''}}

y^{^{''}} - y^{^{'}} - 2 y = 3 e^{(2 x)} - e^{(- x)}

D (r) = \frac{27 ( r ) + 13}{26}

y^{^{''}} + 16 y = 4 (u \frac{7}{4} π)

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = - 20 t^{2} + 52 t - 6

What is the solution for y^{''}+9y=cos(3t),y(pi/6)=1,y(0)=1 ?
The solution for y^{''}+9y=cos(3t),y(pi/6)=1,y(0)=1 is y=cos(3t)+(1-pi/(36))sin(3t)+(tsin(3t))/6