What is the solution for 5/2 x^{''}+10x^'+20x=200,x(0)=0 ?

The solution for 5/2 x^{''}+10x^'+20x=200,x(0)=0 is x=e^{-2t}(-10cos(2t)+c_{2}sin(2t))+10

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5/2 x^{''}+10x^'+20x=200,x(0)=0

\frac{5}{2} x^{^{''}} + 10 x^{^{'}} + 20 x = 200, x (0) = 0

x = e^{- 2 t} (- 10 cos (2 t) + c_{2} sin (2 t)) + 10

Solution steps

\frac{5}{2} x^{''} (t) + 10 x^{'} (t) + 20 x = 200

Solve linear ODE:

x = e^{- 2 t} (- 10 cos (2 t) + c_{2} sin (2 t)) + 10

x = e^{- 2 t} (- 10 cos (2 t) + c_{2} sin (2 t)) + 10

\frac{1}{8} y^{^{''}} + 8 x + 4 y^{^{'}} = 0, y^{^{'}} (0) = 9, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 1 = 0

y^{^{''}} - 16 y = cos (4 t), y (0) = 0, y^{^{'}} (0) = 2

2 y^{^{''}} - y^{^{'}} - 3 y = 3 x e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = \frac{1}{e ^{x} sin ( x )}

What is the solution for 5/2 x^{''}+10x^'+20x=200,x(0)=0 ?
The solution for 5/2 x^{''}+10x^'+20x=200,x(0)=0 is x=e^{-2t}(-10cos(2t)+c_{2}sin(2t))+10