What is the solution for (d^2x)/(dt^2)+16x=1+2sin(4t) ?

The solution for (d^2x)/(dt^2)+16x=1+2sin(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)+1/16-(tcos(4t))/4

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(d^2x)/(dt^2)+16x=1+2sin(4t)

\frac{d ^{2} x}{d t ^{2}} + 16 x = 1 + 2 sin (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{16} - \frac{t cos ( 4 t )}{4}

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 16 x = 1 + 2 sin (4 t)

Solve linear ODE:

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{16} - \frac{t cos ( 4 t )}{4}

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{16} - \frac{t cos ( 4 t )}{4}

y^{^{''}} - 4 y = 2 x - sin (x)

y^{^{''}} + y^{^{'}} - 2 y = 5

\frac{d ^{2} y}{d x ^{2}} + 9 y = x e^{2 x} + sin (3 x)

y^{^{''}} - y^{^{'}} - 12 y = 5 - 2 x

2 \frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = H (t)

What is the solution for (d^2x)/(dt^2)+16x=1+2sin(4t) ?
The solution for (d^2x)/(dt^2)+16x=1+2sin(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)+1/16-(tcos(4t))/4