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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{2 x}{x ^{2} + 4} d x

\frac{d}{d x} (sin ((π x)^{3}))

(2 x + 1)^{4}

x \to \infty lim (lo g_{2} (\frac{( x + 2 )}{( 2 \cdot x + 1 )}))

\int_{1}^{8} x - 1 d x

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