What is the solution for x^{''}+25x=102cos(t) ?

The solution for x^{''}+25x=102cos(t) is x=c_{1}cos(5t)+c_{2}sin(5t)+17/4 cos(t)

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x^{''}+25x=102cos(t)

x^{^{''}} + 25 x = 102 cos (t)

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{17}{4} cos (t)

Solution steps

x^{''} (t) + 25 x = 102 cos (t)

Solve linear ODE:

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{17}{4} cos (t)

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{17}{4} cos (t)

y^{^{''}} - y^{^{'}} = 3 - 4 x

y^{^{''}} + 3 y + 2 y = cos (e^{x})

y^{^{''}} + 15 y^{^{'}} + 7 y = sin (x) + cos (x)

(y^{^{''}}) - 2 y^{^{'}} + y = \frac{e ^{x}}{1 + x ^{2}}

y^{^{''}} + 2 y^{^{'}} + 2 y = t^{2}

What is the solution for x^{''}+25x=102cos(t) ?
The solution for x^{''}+25x=102cos(t) is x=c_{1}cos(5t)+c_{2}sin(5t)+17/4 cos(t)