What is the solution for y^{''}+(3-a)y^'-2(a-1)y=0 ?

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

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y^{''}+(3-a)y^'-2(a-1)y=0

y^{^{''}} + (3 - a) y^{^{'}} - 2 (a - 1) y = 0

y = c_{1} e^{(a - 1) t} + c_{2} e^{- 2 t}

Solution steps

y^{''} (t) + (3 - a) y^{'} (t) - 2 (a - 1) y = 0

Solve linear ODE:

y = c_{1} e^{(a - 1) t} + c_{2} e^{- 2 t}

y = c_{1} e^{(a - 1) t} + c_{2} e^{- 2 t}

- c x^{^{''}} = b x, x (0) = 0

y^{^{''}} - 2 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 4

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (4) = 2, y^{^{'}} (4) = 6

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0

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

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