What is the solution for 6y^{''}+40y^'+54y=0 ?

The solution for 6y^{''}+40y^'+54y=0 is y=c_{1}e^{((-10+sqrt(19))t)/3}+c_{2}e^{-((10+sqrt(19))t)/3}

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6y^{''}+40y^'+54y=0

6 y^{^{''}} + 40 y^{^{'}} + 54 y = 0

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

Solution steps

6 y^{''} (t) + 40 y^{'} (t) + 54 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

y = c_{1} e^{\frac{( - 10 + 19 ) t}{3}} + c_{2} e^{- \frac{( 10 + 19 ) t}{3}}

What is the solution for 6y^{''}+40y^'+54y=0 ?
The solution for 6y^{''}+40y^'+54y=0 is y=c_{1}e^{((-10+sqrt(19))t)/3}+c_{2}e^{-((10+sqrt(19))t)/3}