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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int t (t^{2} + 1)^{3} d t

x \to 10 lim (2^{\frac{1}{x - 10}})

y^{^{'}} = \frac{y ^{2}}{1 + y ^{2}}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

\frac{\partial}{\partial x} (c y^{- 3})

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