What is the solution for 10y^{''}+4y^'-2y=0 ?

The solution for 10y^{''}+4y^'-2y=0 is y=c_{1}e^{((-1+sqrt(6))t)/5}+c_{2}e^{-((1+sqrt(6))t)/5}

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10y^{''}+4y^'-2y=0

10 y^{^{''}} + 4 y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{( - 1 + 6 ) t}{5}} + c_{2} e^{- \frac{( 1 + 6 ) t}{5}}

Solution steps

10 y^{''} (t) + 4 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 6 ) t}{5}} + c_{2} e^{- \frac{( 1 + 6 ) t}{5}}

y = c_{1} e^{\frac{( - 1 + 6 ) t}{5}} + c_{2} e^{- \frac{( 1 + 6 ) t}{5}}

15 y^{^{''''}} + 68 y^{^{''}} + 64 y = 0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0

y^{^{''''}} - 2 y^{^{'''}} - 4 y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 4 y = 0

2 y^{^{''}} + (a - 1) y = 0

What is the solution for 10y^{''}+4y^'-2y=0 ?
The solution for 10y^{''}+4y^'-2y=0 is y=c_{1}e^{((-1+sqrt(6))t)/5}+c_{2}e^{-((1+sqrt(6))t)/5}