What is the solution for 1/2 y^{''}+10y^'+100y=150 ?

The solution for 1/2 y^{''}+10y^'+100y=150 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2

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1/2 y^{''}+10y^'+100y=150

\frac{1}{2} y^{^{''}} + 10 y^{^{'}} + 100 y = 150

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

Solution steps

\frac{1}{2} y^{''} (t) + 10 y^{'} (t) + 100 y = 150

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{4 t}

y^{^{''}} + 2 y^{^{'}} + 10 y = e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} + 5 y = - 20 e^{x} sin (2 x)

2 y^{^{''}} - 2 y^{^{'}} + y = 1 + x + x^{2}

y^{^{''}} + y = 4 cos (x) - sin (x)

What is the solution for 1/2 y^{''}+10y^'+100y=150 ?
The solution for 1/2 y^{''}+10y^'+100y=150 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2