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

The solution for y^{''}+9y= 3/2 sin(2t),y(0)=0,y^'(0)=0 is y=-1/5 sin(3t)+3/10 sin(2t)

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y^{''}+9y= 3/2 sin(2t),y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = \frac{3}{2} sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{5} sin (3 t) + \frac{3}{10} sin (2 t)

Solution steps

y^{''} (t) + 9 y = \frac{3}{2} sin (2 t)

Solve linear ODE:

y = - \frac{1}{5} sin (3 t) + \frac{3}{10} sin (2 t)

y = - \frac{1}{5} sin (3 t) + \frac{3}{10} sin (2 t)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 8

y^{^{''}} - 3 y^{^{'}} - 4 y = 5 e^{- x}, y (0) = 1, y^{^{'}} (0) = 13

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} - 6

y^{^{''}} - y^{^{'}} - 2 y = 1 + t, y (0) = 2, y^{^{'}} (0) = 0

\frac{5}{3} y^{^{''}} + 10 y^{^{'}} + 30 y = 300

What is the solution for y^{''}+9y= 3/2 sin(2t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y= 3/2 sin(2t),y(0)=0,y^'(0)=0 is y=-1/5 sin(3t)+3/10 sin(2t)