What is the solution for y^{''}+20y^'+200y=300 ?

The solution for y^{''}+20y^'+200y=300 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2

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y^{''}+20y^'+200y=300

y^{^{''}} + 20 y^{^{'}} + 200 y = 300

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

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 y = 300

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{d}{d x} (x^{3} - 27 x)

\int sin (ln (t)) d t

\int \frac{sin ( 1 + x )}{1 + x} d x

\frac{d}{d x} (sin (\frac{1 - e ^{2 x}}{1 + e ^{2 x}}))

\int 2 t + 3 d t

What is the solution for y^{''}+20y^'+200y=300 ?
The solution for y^{''}+20y^'+200y=300 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2