What is the solution for 6y^{''}+y^'-2y=0,y(1)=2,y^'(1)=-1/6 ?

The solution for 6y^{''}+y^'-2y=0,y(1)=2,y^'(1)=-1/6 is y=e^{(3t-7)/6+2/3}+e^{(2-2t)/3}

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6y^{''}+y^'-2y=0,y(1)=2,y^'(1)=-1/6

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = - \frac{1}{6}

y = e^{\frac{3 t - 7}{6} + \frac{2}{3}} + e^{\frac{2 - 2 t}{3}}

Solution steps

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

Solve linear ODE:

y = e^{\frac{3 t - 7}{6} + \frac{2}{3}} + e^{\frac{2 - 2 t}{3}}

y = e^{\frac{3 t - 7}{6} + \frac{2}{3}} + e^{\frac{2 - 2 t}{3}}

y^{^{''}} - 2 \cdot y^{^{'}} + y = 0

25 y^{^{''}} + 40 y^{^{'}} + 16 y = 0

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} - 8 y + 17 y = 0

What is the solution for 6y^{''}+y^'-2y=0,y(1)=2,y^'(1)=-1/6 ?
The solution for 6y^{''}+y^'-2y=0,y(1)=2,y^'(1)=-1/6 is y=e^{(3t-7)/6+2/3}+e^{(2-2t)/3}