What is the solution for y^{''}+9y=cos(3t),y(0)=1,y^'(0)=2 ?

The solution for y^{''}+9y=cos(3t),y(0)=1,y^'(0)=2 is y=cos(3t)+2/3 sin(3t)+(tsin(3t))/6

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y^{''}+9y=cos(3t),y(0)=1,y^'(0)=2

y^{^{''}} + 9 y = cos (3 t), y (0) = 1, y^{^{'}} (0) = 2

y = cos (3 t) + \frac{2}{3} 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) + \frac{2}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y = cos (3 t) + \frac{2}{3} sin (3 t) + \frac{t sin ( 3 t )}{6}

y^{^{''}} - 4 y^{^{'}} - 5 y = 12, 5 y (0) = 0

y^{^{''}} - y^{^{'}} + y = 6 sin (x) - 3 e^{2 x}

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} + x = 10 sin (4 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = cos (2 t)

y^{^{''}} + 10 y^{^{'}} + 24 y = 2 e^{- 4 x} (4 x - 2)

What is the solution for y^{''}+9y=cos(3t),y(0)=1,y^'(0)=2 ?
The solution for y^{''}+9y=cos(3t),y(0)=1,y^'(0)=2 is y=cos(3t)+2/3 sin(3t)+(tsin(3t))/6