What is the solution for (t^2-10t+21)y^'=y ?

The solution for (t^2-10t+21)y^'=y is y=(c_{1}\sqrt[4]{t-7})/(\sqrt[4]{t-3)}

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(t^2-10t+21)y^'=y

(t^{2} - 10 t + 21) y^{^{'}} = y

Solution steps

(t^{2} - 10 t + 21) y^{'} (t) = y

Solve separable ODE:

\frac{d y}{d x} = 10 x^{9} y

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

\frac{x ^{2} + 2 x}{x + 1} y^{^{'}} - \frac{1}{y ^{3}} = 0

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

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

What is the solution for (t^2-10t+21)y^'=y ?
The solution for (t^2-10t+21)y^'=y is y=(c_{1}\sqrt[4]{t-7})/(\sqrt[4]{t-3)}