What is the solution for 0.5y^{''}+50y=cos(t) ?

The solution for 0.5y^{''}+50y=cos(t) is y=c_{1}cos(10t)+c_{2}sin(10t)+2/99 cos(t)

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0.5y^{''}+50y=cos(t)

0.5 y^{^{''}} + 50 y = cos (t)

y = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{2}{99} cos (t)

Solution steps

0.5 y^{''} (t) + 50 y = cos (t)

Solve linear ODE:

y = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{2}{99} cos (t)

y = c_{1} cos (10 t) + c_{2} sin (10 t) + \frac{2}{99} cos (t)

x^{^{''}} + x = 1, x (0) = 2, x^{^{'}} (0) = - 1

4 y^{^{''}} + y = e^{x} x^{2}

\frac{d ^{2} y}{d x ^{2}} - 4 y^{^{'}} + 3 y = 8 e^{- x} + 9

(d^{2} + 9) y = cos^{4} (x)

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

What is the solution for 0.5y^{''}+50y=cos(t) ?
The solution for 0.5y^{''}+50y=cos(t) is y=c_{1}cos(10t)+c_{2}sin(10t)+2/99 cos(t)