What is the solution for 0.2q^{''}+4q^'+100q=0 ?

The solution for 0.2q^{''}+4q^'+100q=0 is q(t)=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))

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0.2q^{''}+4q^'+100q=0

0.2 q^{^{''}} + 4 q^{^{'}} + 100 q = 0

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

Solution steps

0.2 q^{''} (t) + 4 q^{'} (t) + 100 q (t) = 0

Solve linear ODE:

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

y^{^{''}} - 20 y^{^{'}} + 100 y = 0, y (0) = 4, y (1) = 3

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 8 y^{^{'}} - 2 y = 0

y^{^{''}} + 4 y = 0, y (0) = y^{^{'}} (0)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 3 y = 0

What is the solution for 0.2q^{''}+4q^'+100q=0 ?
The solution for 0.2q^{''}+4q^'+100q=0 is q(t)=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))