What is the solution for x^{''}+12x^'+40x=0,x(0)=0,x^'(0)=4 ?

The solution for x^{''}+12x^'+40x=0,x(0)=0,x^'(0)=4 is x=2e^{-6t}sin(2t)

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x^{''}+12x^'+40x=0,x(0)=0,x^'(0)=4

x^{^{''}} + 12 x^{^{'}} + 40 x = 0, x (0) = 0, x^{^{'}} (0) = 4

x = 2 e^{- 6 t} sin (2 t)

Solution steps

x^{''} (t) + 12 x^{'} (t) + 40 x = 0

Solve linear ODE:

x = 2 e^{- 6 t} sin (2 t)

x = 2 e^{- 6 t} sin (2 t)

13 y + 6 y^{^{'}} + y^{^{''}} = 0

y^{^{'''}} + (- 3) y^{^{''}} + 3 y^{^{'}} - y = 0

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

3 x^{^{''}} + 75 x = 0, x (0) = 0, x^{^{'}} (0) = - 10

12 y^{^{''}} - 13 y^{^{'}} + y = 0, y (0) = 12, y^{^{'}} (0) = \frac{1}{12}

What is the solution for x^{''}+12x^'+40x=0,x(0)=0,x^'(0)=4 ?
The solution for x^{''}+12x^'+40x=0,x(0)=0,x^'(0)=4 is x=2e^{-6t}sin(2t)