What is the solution for (t^2-22t+96)(dy)/(dt)=y ?

The solution for (t^2-22t+96)(dy)/(dt)=y is y=(c_{1}\sqrt[10]{t-16})/(\sqrt[10]{t-6)}

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(t^2-22t+96)(dy)/(dt)=y

(t^{2} - 22 t + 96) \frac{d y}{d t} = y

Solution steps

(t^{2} - 22 t + 96) \frac{d y}{d t} = y

Solve separable ODE:

\frac{d y}{d x} = \frac{x sin ( x )}{y}

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

e^{3 x} y - e^{5 x} y^{3} y^{^{'}} = 0

x \cdot y^{^{'}} + y = 0

y^{^{'}} = y - y^{2}, y (- 1) = 2

What is the solution for (t^2-22t+96)(dy)/(dt)=y ?
The solution for (t^2-22t+96)(dy)/(dt)=y is y=(c_{1}\sqrt[10]{t-16})/(\sqrt[10]{t-6)}