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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 e^{x} - 18 e^{- x}

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

y^{^{''}} - 2 y^{^{'}} + y = e^{2 t}

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

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