What is the solution for 24y^{''}+7y^'-5y=0 ?

The solution for 24y^{''}+7y^'-5y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(5t)/8}

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24y^{''}+7y^'-5y=0

24 y^{^{''}} + 7 y^{^{'}} - 5 y = 0

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{5 t}{8}}

Solution steps

24 y^{''} (t) + 7 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{5 t}{8}}

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{5 t}{8}}

2 y^{^{''}} + 12 y^{^{'}} + 18 y = 0

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

x^{^{''}} + 42 x^{^{'}} + 8 x = 0

y^{^{''}} + 2 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for 24y^{''}+7y^'-5y=0 ?
The solution for 24y^{''}+7y^'-5y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(5t)/8}