What is the solution for x^{''}+3x^'+4x=0 ?

The solution for x^{''}+3x^'+4x=0 is x=e^{-(3t)/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

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x^{''}+3x^'+4x=0

x^{^{''}} + 3 x^{^{'}} + 4 x = 0

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

x^{''} (t) + 3 x^{'} (t) + 4 x = 0

Solve linear ODE:

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

x = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y^{^{''}} + 4 y + 4 y = 0

y^{^{''}} - 10 y^{^{'}} + 16 y = 0

y^{^{''}} - 19 y^{^{'}} - 10 y = 0

y^{^{''}} + y = 0, y (\frac{π}{2}) = 5, y (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 6

What is the solution for x^{''}+3x^'+4x=0 ?
The solution for x^{''}+3x^'+4x=0 is x=e^{-(3t)/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))