What is the solution for y^{''}+20y^'+200y=300,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+20y^'+200y=300,y(0)=0,y^'(0)=0 is y=e^{-10t}(-3/2 cos(10t)-3/2 sin(10t))+3/2

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y^{''}+20y^'+200y=300,y(0)=0,y^'(0)=0

y^{^{''}} + 20 y^{^{'}} + 200 y = 300, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = 9 + 8 x + 6 x e^{x}, at x = 0

\frac{\partial}{\partial x} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

t \to 1 lim (\frac{2 t}{t ^{2} - 1})

n = 1 \sum \infty 3 (5)^{- n}

x \to 0 lim (9 x - 8 x)

What is the solution for y^{''}+20y^'+200y=300,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+20y^'+200y=300,y(0)=0,y^'(0)=0 is y=e^{-10t}(-3/2 cos(10t)-3/2 sin(10t))+3/2