What is the solution for 3y^{''}+8y^'-16y=0 ?

The solution for 3y^{''}+8y^'-16y=0 is y=c_{1}e^{(4t)/3}+c_{2}e^{-4t}

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3y^{''}+8y^'-16y=0

3 y^{^{''}} + 8 y^{^{'}} - 16 y = 0

y = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{- 4 t}

Solution steps

3 y^{''} (t) + 8 y^{'} (t) - 16 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{- 4 t}

y = c_{1} e^{\frac{4 t}{3}} + c_{2} e^{- 4 t}

y^{^{''}} + 4 y = e^{- x}

\frac{\partial}{\partial x} (t^{2} e^{- x})

\int 2 x (x^{2} + 1)^{5} d x

\frac{\partial}{\partial x} (ln (5 x))

\frac{d}{d x} ((x^{3} - 4) arcsin (x))

What is the solution for 3y^{''}+8y^'-16y=0 ?
The solution for 3y^{''}+8y^'-16y=0 is y=c_{1}e^{(4t)/3}+c_{2}e^{-4t}