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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (11 + 3 s + 5 y, 2 + 5 + 2 y, 3 + 2 x + y)

x \to 0 lim ({x})

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\int \frac{2 x + 3}{( x - 1 ) ^{2}} d x

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