What is the solution for 3y^{''}+3y^'-2y=0 ?

The solution for 3y^{''}+3y^'-2y=0 is y=c_{1}e^{((-3+sqrt(33))t)/6}+c_{2}e^{((-3-sqrt(33))t)/6}

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3y^{''}+3y^'-2y=0

3 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{( - 3 + 33 ) t}{6}} + c_{2} e^{\frac{( - 3 - 33 ) t}{6}}

Solution steps

3 y^{''} (t) + 3 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 33 ) t}{6}} + c_{2} e^{\frac{( - 3 - 33 ) t}{6}}

y = c_{1} e^{\frac{( - 3 + 33 ) t}{6}} + c_{2} e^{\frac{( - 3 - 33 ) t}{6}}

e^{(x^{2})}

x \to 1 + lim (\frac{2 x + 1}{x - 1})

y^{^{''}} + 12 y^{^{'}} + 37 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{1} \frac{23}{4 y - 1} d y

f (x) = cos^{3} (π x)

What is the solution for 3y^{''}+3y^'-2y=0 ?
The solution for 3y^{''}+3y^'-2y=0 is y=c_{1}e^{((-3+sqrt(33))t)/6}+c_{2}e^{((-3-sqrt(33))t)/6}