What is the solution for (t^2-24t+95)y^'=y ?

The solution for (t^2-24t+95)y^'=y is y=(c_{1}\sqrt[14]{t-19})/(\sqrt[14]{t-5)}

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(t^2-24t+95)y^'=y

(t^{2} - 24 t + 95) y^{^{'}} = y

Solution steps

(t^{2} - 24 t + 95) y^{'} (t) = y

Solve separable ODE:

\frac{d}{d x} (x^{4} + y^{4}) = \frac{d}{d x} (16)

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

\frac{d y}{d t} = k y + n

x \frac{d y}{d x} = e^{2 y + x}

y^{^{'}} = \frac{( 5 x ^{2} - 1 )}{5 + 4 y}

What is the solution for (t^2-24t+95)y^'=y ?
The solution for (t^2-24t+95)y^'=y is y=(c_{1}\sqrt[14]{t-19})/(\sqrt[14]{t-5)}