What is the solution for y^{''}+16y=3.36 ?

The solution for y^{''}+16y=3.36 is y=c_{1}cos(4t)+c_{2}sin(4t)+0.21

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y^{''}+16y=3.36

y^{^{''}} + 16 y = 3.36

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 0.21

Solution steps

y^{''} (t) + 16 y = 3.36

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 0.21

y = c_{1} cos (4 t) + c_{2} sin (4 t) + 0.21

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (x)

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 4 y = 11 t^{3} sin (2 t)

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

What is the solution for y^{''}+16y=3.36 ?
The solution for y^{''}+16y=3.36 is y=c_{1}cos(4t)+c_{2}sin(4t)+0.21