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

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

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

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

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

Solution steps

y^{''} (t) + y = sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} tan^{2} (x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 e^{- x}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 7) e^{- 2 x}

- y^{^{'''}} - y^{^{'}} = 2 cos (t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + 6

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