What is the solution for (t^2-26t+120)y^'=y ?

The solution for (t^2-26t+120)y^'=y is y=(c_{1}\sqrt[14]{t-20})/(\sqrt[14]{t-6)}

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(t^2-26t+120)y^'=y

(t^{2} - 26 t + 120) y^{^{'}} = y

Solution steps

(t^{2} - 26 t + 120) y^{'} (t) = y

Solve separable ODE:

y^{^{'}} = \frac{( y - 2 ) ^{2}}{x ^{2} + 1}

\frac{d y}{y} = 34 y x^{16} d x

y^{^{'}} + \frac{y}{8 x} = 0, y (5) = e

\frac{( x + a ) d y}{d x - y} = b

e^{- y} cos (x) - \frac{d y}{d x} = 0

What is the solution for (t^2-26t+120)y^'=y ?
The solution for (t^2-26t+120)y^'=y is y=(c_{1}\sqrt[14]{t-20})/(\sqrt[14]{t-6)}