What is the solution for 3y^{''}+5y^'-y=0 ?

The solution for 3y^{''}+5y^'-y=0 is y=c_{1}e^{((-5+sqrt(37))t)/6}+c_{2}e^{((-5-sqrt(37))t)/6}

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3y^{''}+5y^'-y=0

3 y^{^{''}} + 5 y^{^{'}} - y = 0

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}}

Solution steps

3 y^{''} (t) + 5 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}}

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 10 y = 0

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

2 x^{^{''}} + 10 x = 0

2 y^{^{'''}} - 6 y^{^{''}} = 0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = \frac{1}{3}

What is the solution for 3y^{''}+5y^'-y=0 ?
The solution for 3y^{''}+5y^'-y=0 is y=c_{1}e^{((-5+sqrt(37))t)/6}+c_{2}e^{((-5-sqrt(37))t)/6}