What is the solution for y^{''}+25y= 2/3 cos(2)t ?

The solution for y^{''}+25y= 2/3 cos(2)t is y=c_{1}cos(5t)+c_{2}sin(5t)+(2cos(2)t)/(75)

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y^{''}+25y= 2/3 cos(2)t

y^{^{''}} + 25 y = \frac{2}{3} cos (2) t

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{2 cos ( 2 ) t}{75}

Solution steps

y^{''} (t) + 25 y = \frac{2}{3} cos (2) t

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{2 cos ( 2 ) t}{75}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{2 cos ( 2 ) t}{75}

3 y^{^{''}} + y^{^{'}} = 9 x

x^{^{''}} + 5 x^{^{'}} + 6 x = 2 e^{- t}, x (0) = 1, x^{^{'}} (0) = 0

y^{^{''}} - y = 2 sin (x)

x^{^{''}} - x = 4 e^{t}

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

What is the solution for y^{''}+25y= 2/3 cos(2)t ?
The solution for y^{''}+25y= 2/3 cos(2)t is y=c_{1}cos(5t)+c_{2}sin(5t)+(2cos(2)t)/(75)