What is the solution for y^{''}+16y=2sin(2t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+16y=2sin(2t),y(0)=1,y^'(0)=0 is y=cos(4t)-1/12 sin(4t)+1/6 sin(2t)

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y^{''}+16y=2sin(2t),y(0)=1,y^'(0)=0

y^{^{''}} + 16 y = 2 sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

y = cos (4 t) - \frac{1}{12} sin (4 t) + \frac{1}{6} sin (2 t)

Solution steps

y^{''} (t) + 16 y = 2 sin (2 t)

Solve linear ODE:

y = cos (4 t) - \frac{1}{12} sin (4 t) + \frac{1}{6} sin (2 t)

y = cos (4 t) - \frac{1}{12} sin (4 t) + \frac{1}{6} sin (2 t)

\int (2 t e^{t}) d t

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

y = e^{4 x} cos (7 x)

4 x ln (x)

\frac{\partial}{\partial y} (5 x^{2} - 3 x y^{2} + x yz)

What is the solution for y^{''}+16y=2sin(2t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+16y=2sin(2t),y(0)=1,y^'(0)=0 is y=cos(4t)-1/12 sin(4t)+1/6 sin(2t)