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

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

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y^{''}-(2α-3)y^'+α(α-3)y=0

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

y = c_{1} e^{α t} + c_{2} e^{(α - 3) t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{α t} + c_{2} e^{(α - 3) t}

y = c_{1} e^{α t} + c_{2} e^{(α - 3) t}

y^{^{''}} + 9 y^{^{'}} + 12 y = 0

3 y^{^{''}} - 16 y^{^{'}} + 13 y = 0

y^{^{''}} + n^{2} y = 0, y (0) = c, y^{^{'}} (0) = v

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = \frac{1}{2}

24 y^{^{''}} + 2.2 y = 0

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