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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

n = 1 \sum \infty \frac{7 ^{n}}{6 ^{n}}

x \to \infty lim (5 e^{- t} i - \frac{5 t}{t + 7} j + 5 arctan (t k))

y^{^{'}} = - 6 y

\frac{\partial}{\partial x} (x^{2} + y^{2} + 3 x y + 5 x)

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