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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{1}^{b} \frac{2 x}{x ^{2} + 4} d x

x^{2} y + y^{^{'}} = - 3 x^{2}

f (x) = 1 + 4 x, at x = 2

x \to c lim (x^{3} + 4 x^{2} - 3)

\int_{0}^{1} (3 + 2 x^{2}) d x

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