What is the solution for 3y^{''}+17y^'-6y=0 ?

The solution for 3y^{''}+17y^'-6y=0 is y=c_{1}e^{t/3}+c_{2}e^{-6t}

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3y^{''}+17y^'-6y=0

3 y^{^{''}} + 17 y^{^{'}} - 6 y = 0

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- 6 t}

Solution steps

3 y^{''} (t) + 17 y^{'} (t) - 6 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- 6 t}

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- 6 t}

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

\int (x^{2} - 5)^{5} \cdot 2 x d x

y^{^{''}} = x e^{x}

f (x) = x^{3} - 2 x^{2} + 4 x + 3, at x = 1

10 x - 8 y \cdot x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 4

What is the solution for 3y^{''}+17y^'-6y=0 ?
The solution for 3y^{''}+17y^'-6y=0 is y=c_{1}e^{t/3}+c_{2}e^{-6t}