What is the solution for y^{''}+16y=16sin(4t) ?

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

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y^{''}+16y=16sin(4t)

y^{^{''}} + 16 y = 16 sin (4 t)

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

Solution steps

y^{''} (t) + 16 y = 16 sin (4 t)

Solve linear ODE:

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

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

\int \frac{1}{cos ^{2} ( 2 y )} d y

f (x) = 14 (2 x + 5)^{6}

\frac{\partial}{\partial y} (\frac{y ^{3}}{x})

t^{8} cos (t)

t^{2} \cdot e^{-, at}

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