What is the solution for y^{''}+16y=2sin(4t) ?

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

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y^{''}+16y=2sin(4t)

y^{^{''}} + 16 y = 2 sin (4 t)

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

Solution steps

y^{''} (t) + 16 y = 2 sin (4 t)

Solve linear ODE:

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

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

x \to - 3 lim (- 2)

x \to \infty lim (e^{\frac{1}{x}} sin (\frac{1}{x}))

y = (\frac{x ^{2} + 2}{x ^{2} - 2})^{8}

\int cot (a x) d x

\int csc^{2} (x) cot^{5} (x) d x

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