What is the solution for (t^2-16t+15)y^'=y ?

The solution for (t^2-16t+15)y^'=y is y=(c_{1}\sqrt[14]{t-15})/(\sqrt[14]{t-1)}

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(t^2-16t+15)y^'=y

(t^{2} - 16 t + 15) y^{^{'}} = y

Solution steps

(t^{2} - 16 t + 15) y^{'} (t) = y

Solve separable ODE:

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = \frac{- 1}{6}

\frac{d y}{d x} - y^{5} = 0, y (0) = 1

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

(1 + x^{2}) \frac{d y}{d x} + 2 x y = - x, x > 0

(x + 2) \frac{d y}{d x} = y

What is the solution for (t^2-16t+15)y^'=y ?
The solution for (t^2-16t+15)y^'=y is y=(c_{1}\sqrt[14]{t-15})/(\sqrt[14]{t-1)}