What is the solution for 0.5y^{''}+10y^'+100y=150 ?

The solution for 0.5y^{''}+10y^'+100y=150 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2

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0.5y^{''}+10y^'+100y=150

0.5 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

0.5 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}

x \to 1 lim ((x)^{\frac{1}{1 - x}})

\frac{d}{d x} (- \frac{2 x}{y})

\int \frac{4 x ^{2} - 25}{x} d x

(- 6, 1), (- 4, - 9)

\frac{d}{d x} (4 x + 2)

What is the solution for 0.5y^{''}+10y^'+100y=150 ?
The solution for 0.5y^{''}+10y^'+100y=150 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2