What is the solution for 6y^{''}+y^'-2y=0 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2}}{d x ^{2}} (e^{- k x^{2}})

\int 2 x 9 - x^{2} d x

x y^{^{'}} = y + 2 x^{2} y, y (1) = 1

g (x) = 5 tan (3 x)

\int x sin^{2} (8 x) d x

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