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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{9}{( 7 x + 4 ) ^{5}} d x

\int 2 x^{2} cos (5 x) d x

y^{^{'}} + 7 y = 5 cos (2 t)

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

x \to 0 + lim (2^{- \frac{1}{x}})

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