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

The solution for y^{''}+y=2sin(2t),y(0)=9,y^'(0)=0 is y=9cos(t)+4/3 sin(t)-2/3 sin(2t)

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

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

y = 9 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 = 9 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

y = 9 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

y^{^{''}} + 4 y = sec (x)

y^{^{''}} + 2 y^{^{'}} + y = \frac{( e ^{x} )}{x}

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

y^{^{''}} + 5 y^{^{'}} = 2 x^{3} - 4 x^{2} - x + 6

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 \frac{e ^{- 4 x}}{x ^{4}}

What is the solution for y^{''}+y=2sin(2t),y(0)=9,y^'(0)=0 ?
The solution for y^{''}+y=2sin(2t),y(0)=9,y^'(0)=0 is y=9cos(t)+4/3 sin(t)-2/3 sin(2t)