What is the solution for x^{''}+2x^'+0.75x=0 ?

The solution for x^{''}+2x^'+0.75x=0 is x=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}

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x^{''}+2x^'+0.75x=0

x^{^{''}} + 2 x^{^{'}} + 0.75 x = 0

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

Solution steps

x^{''} (t) + 2 x^{'} (t) + 0.75 x = 0

Solve linear ODE:

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y^{^{''}} = 7 y^{^{'}} - 12 y, y (0) = 0, y^{^{'}} (0) = - 1

u^{^{''}} + 5 H (t) = 0

y^{^{''}} - 10 y^{^{'}} + 2 y = 0, y (0) = 0

\frac{d ^{4} y}{d x ^{4}} - 2 y^{^{''}} + 1 y = 0

u^{^{''}} + H (t) = 0

What is the solution for x^{''}+2x^'+0.75x=0 ?
The solution for x^{''}+2x^'+0.75x=0 is x=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}