What is the solution for y^{''}+4y=20sin(2t) ?

The solution for y^{''}+4y=20sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-5tcos(2t)

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y^{''}+4y=20sin(2t)

y^{^{''}} + 4 y = 20 sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 5 t cos (2 t)

Solution steps

y^{''} (t) + 4 y = 20 sin (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 5 t cos (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 5 t cos (2 t)

f (x) = \frac{5 x}{x + 3}, (2, 2)

13.375 ln (\frac{x}{x - 1250})

\int 6 x (1 - x) d x

t^{2} \cdot y^{^{'}} + t y = 7, y (1) = 7

\int 8 sin^{2} (x) cos (x) d x

What is the solution for y^{''}+4y=20sin(2t) ?
The solution for y^{''}+4y=20sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-5tcos(2t)