What is the solution for y^{''}+y=2sin(2t),y(0)=6,y^'(0)=0 ?

The solution for y^{''}+y=2sin(2t),y(0)=6,y^'(0)=0 is y=6cos(t)+4/3 sin(t)-2/3 sin(2t)

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y^{''}+y=2sin(2t),y(0)=6,y^'(0)=0

y^{^{''}} + y = 2 sin (2 t), y (0) = 6, y^{^{'}} (0) = 0

y = 6 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 6 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

y = 6 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

d, d (t) = 12 t + 20

y^{^{''}} + 2 y^{^{'}} + y = 12 x e^{- x}

2 y^{^{''}} + 5 y^{^{'}} + 4 x = 0

y^{^{''}} - y^{^{'}} = 3 x^{2}

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 3 + \frac{e ^{x}}{x}

What is the solution for y^{''}+y=2sin(2t),y(0)=6,y^'(0)=0 ?
The solution for y^{''}+y=2sin(2t),y(0)=6,y^'(0)=0 is y=6cos(t)+4/3 sin(t)-2/3 sin(2t)