What is the solution for y^{''}-(2α-12)y^'+(α^2-12α+32)y=0 ?

The solution for y^{''}-(2α-12)y^'+(α^2-12α+32)y=0 is y=c_{1}e^{(α-4)t}+c_{2}e^{(α-8)t}

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y^{''}-(2α-12)y^'+(α^2-12α+32)y=0

y^{^{''}} - (2 α - 12) y^{^{'}} + (α^{2} - 12 α + 32) y = 0

y = c_{1} e^{(α - 4) t} + c_{2} e^{(α - 8) t}

Solution steps

y^{''} (t) - (2 α - 12) y^{'} (t) + (α^{2} - 12 α + 32) y = 0

Solve linear ODE:

y = c_{1} e^{(α - 4) t} + c_{2} e^{(α - 8) t}

y = c_{1} e^{(α - 4) t} + c_{2} e^{(α - 8) t}

3 y^{^{''}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 12

y = \frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x}

36 y^{^{''}} - 81 y = 0, y (0) = 0, y^{^{'}} (0) = 1

4 x^{^{''}} + 20 x^{^{'}} + 169 x = 0, x (0) = 3, x^{^{'}} (0) = 12

\frac{4}{32} y^{^{''}} + 3 y = 0

What is the solution for y^{''}-(2α-12)y^'+(α^2-12α+32)y=0 ?
The solution for y^{''}-(2α-12)y^'+(α^2-12α+32)y=0 is y=c_{1}e^{(α-4)t}+c_{2}e^{(α-8)t}