What is the solution for (t^2-22t+96)y^'=y ?

The solution for (t^2-22t+96)y^'=y is y=(c_{1}\sqrt[10]{t-16})/(\sqrt[10]{t-6)}

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(t^2-22t+96)y^'=y

(t^{2} - 22 t + 96) y^{^{'}} = y

Solution steps

(t^{2} - 22 t + 96) y^{'} (t) = y

Solve separable ODE:

5 d x - e^{y - x} d y = 0

\frac{d y}{d x} = 3 x^{2} y, y = 9, x = 0

\frac{d y}{d x} = 3 y + 2, y (0) = 1

y^{^{'}} = 5 y, y (0) = 10

\frac{d y}{d x} = (\frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )})

What is the solution for (t^2-22t+96)y^'=y ?
The solution for (t^2-22t+96)y^'=y is y=(c_{1}\sqrt[10]{t-16})/(\sqrt[10]{t-6)}