What is the solution for y^{''}+y=2sin(2t),y(0)=12,y^'(0)=0 ?

The solution for y^{''}+y=2sin(2t),y(0)=12,y^'(0)=0 is y=12cos(t)+4/3 sin(t)-2/3 sin(2t)

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y^{''}+y=2sin(2t),y(0)=12,y^'(0)=0

y^{^{''}} + y = 2 sin (2 t), y (0) = 12, y^{^{'}} (0) = 0

y = 12 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 = 12 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

y = 12 cos (t) + \frac{4}{3} sin (t) - \frac{2}{3} sin (2 t)

10 y^{^{''}} + 90 y^{^{'}} + \frac{1}{0.05} y = 500 sin (x)

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

y^{^{''}} + 3 y^{^{'}} - 1 = 2 x^{2} + x

y^{^{''}} - 4 y = 24 cos (2 t)

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 cos (x) + cos (3 x)

What is the solution for y^{''}+y=2sin(2t),y(0)=12,y^'(0)=0 ?
The solution for y^{''}+y=2sin(2t),y(0)=12,y^'(0)=0 is y=12cos(t)+4/3 sin(t)-2/3 sin(2t)