What is the solution for 6y^{''}-y^'-35y=0 ?

The solution for 6y^{''}-y^'-35y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{-(7t)/3}

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6y^{''}-y^'-35y=0

6 y^{^{''}} - y^{^{'}} - 35 y = 0

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{- \frac{7 t}{3}}

Solution steps

6 y^{''} (t) - y^{'} (t) - 35 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{- \frac{7 t}{3}}

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{- \frac{7 t}{3}}

4 y^{^{'''}} + 5 y^{^{''}} - 23 y^{^{'}} - 6 y = 0

x^{^{''}} - 6 x^{^{'}} + 8 x = 0, x (0) = 3

y^{^{''}} + y - 6 y^{^{'}} = 0

y^{^{''}} - 8 y^{^{'}} - 7 y = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (0) = 1

What is the solution for 6y^{''}-y^'-35y=0 ?
The solution for 6y^{''}-y^'-35y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{-(7t)/3}