What is the solution for 10y^{''}+12y^'-18y=0 ?

The solution for 10y^{''}+12y^'-18y=0 is y=c_{1}e^{(3(sqrt(6)-1)t)/5}+c_{2}e^{-(3(1+sqrt(6))t)/5}

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10y^{''}+12y^'-18y=0

10 y^{^{''}} + 12 y^{^{'}} - 18 y = 0

y = c_{1} e^{\frac{3 ( 6 - 1 ) t}{5}} + c_{2} e^{- \frac{3 ( 1 + 6 ) t}{5}}

Solution steps

10 y^{''} (t) + 12 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{3 ( 6 - 1 ) t}{5}} + c_{2} e^{- \frac{3 ( 1 + 6 ) t}{5}}

y = c_{1} e^{\frac{3 ( 6 - 1 ) t}{5}} + c_{2} e^{- \frac{3 ( 1 + 6 ) t}{5}}

y^{^{'''}} + 19 y^{^{''}} + 97 y^{^{'}} + 79 y = 0

y^{^{''}} + 3 y^{^{'}} - 16 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = 2

x^{^{''}} + x^{^{'}} + \frac{1}{2} x = 0

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

What is the solution for 10y^{''}+12y^'-18y=0 ?
The solution for 10y^{''}+12y^'-18y=0 is y=c_{1}e^{(3(sqrt(6)-1)t)/5}+c_{2}e^{-(3(1+sqrt(6))t)/5}