What is the solution for y^{''}+4y^'+2.5y=0 ?

The solution for y^{''}+4y^'+2.5y=0 is y=c_{1}e^{((-4+sqrt(6))t)/2}+c_{2}e^{-((4+sqrt(6))t)/2}

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y^{''}+4y^'+2.5y=0

y^{^{''}} + 4 y^{^{'}} + 2.5 y = 0

y = c_{1} e^{\frac{( - 4 + 6 ) t}{2}} + c_{2} e^{- \frac{( 4 + 6 ) t}{2}}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 2.5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 4 + 6 ) t}{2}} + c_{2} e^{- \frac{( 4 + 6 ) t}{2}}

y = c_{1} e^{\frac{( - 4 + 6 ) t}{2}} + c_{2} e^{- \frac{( 4 + 6 ) t}{2}}

y^{^{'''}} - 3 y^{^{'}} - 2 y = 0

(d^{3} + d^{2} + 4 d + 4) \cdot y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

(d^{3} - 5 d^{2} + 8 d - 4) \cdot y = 0

(d^{3} - d^{2} - d + 1) y = 0

What is the solution for y^{''}+4y^'+2.5y=0 ?
The solution for y^{''}+4y^'+2.5y=0 is y=c_{1}e^{((-4+sqrt(6))t)/2}+c_{2}e^{-((4+sqrt(6))t)/2}