What is the solution for 3y^{''}-14y^'+8y=0 ?

The solution for 3y^{''}-14y^'+8y=0 is y=c_{1}e^{4t}+c_{2}e^{(2t)/3}

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3y^{''}-14y^'+8y=0

3 y^{^{''}} - 14 y^{^{'}} + 8 y = 0

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

Solution steps

3 y^{''} (t) - 14 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

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

\frac{d}{d x} (sec (x) tan (x) x (x + 1))

x^{3} \frac{d y}{d x} + 4 x^{2} y = e^{- x}

\int 5 sin (3 x) sin (6 x) d x

\frac{15}{3 s + 3}

What is the solution for 3y^{''}-14y^'+8y=0 ?
The solution for 3y^{''}-14y^'+8y=0 is y=c_{1}e^{4t}+c_{2}e^{(2t)/3}