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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d y} (x^{2} y sin^{3} (x y^{2}))

x \to 0 lim (arcsin (e^{x}))

\frac{d}{d x} (\frac{( x - 1 )}{x + 1})

\frac{d}{d x} (\frac{5 x ^{4}}{2 - x ^{2}})

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

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