What is the solution for 6y^{''}+23y^'-4y=0 ?

The solution for 6y^{''}+23y^'-4y=0 is y=c_{1}e^{t/6}+c_{2}e^{-4t}

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

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

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- 4 t}

Solution steps

6 y^{''} (t) + 23 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- 4 t}

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- 4 t}

\frac{d}{d x} ((x - 4) (\frac{x}{2} + 1)^{3})

\frac{\partial}{\partial x} (7 e^{x y + 8})

\frac{d}{d x} ((x + 1)^{3} (x - 1)^{2})

\int e^{7 x - 5} d x

3 u

What is the solution for 6y^{''}+23y^'-4y=0 ?
The solution for 6y^{''}+23y^'-4y=0 is y=c_{1}e^{t/6}+c_{2}e^{-4t}