What is the solution for 1/2 y^{''}-1/3 y^'-2y=0 ?

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

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1/2 y^{''}-1/3 y^'-2y=0

\frac{1}{2} y^{^{''}} - \frac{1}{3} y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{( 1 + 37 ) t}{3}} + c_{2} e^{\frac{( 1 - 37 ) t}{3}}

Solution steps

\frac{1}{2} y^{''} (t) - \frac{1}{3} y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 37 ) t}{3}} + c_{2} e^{\frac{( 1 - 37 ) t}{3}}

y = c_{1} e^{\frac{( 1 + 37 ) t}{3}} + c_{2} e^{\frac{( 1 - 37 ) t}{3}}

y^{^{''}} + y^{^{'}} + 9 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + y 3 = 0

y^{^{''}} + 36 y = 0, y (0) = cos (6 x)

5 y^{^{''}} + 30 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

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