What is the solution for (t^2-16t+15)(dy)/(dt)=y ?

The solution for (t^2-16t+15)(dy)/(dt)=y is y=(c_{1}\sqrt[14]{t-15})/(\sqrt[14]{t-1)}

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(t^2-16t+15)(dy)/(dt)=y

(t^{2} - 16 t + 15) \frac{d y}{d t} = y

y = \frac{c _{1} 14 t - 15}{14 t - 1}

Solution steps

(t^{2} - 16 t + 15) \frac{d y}{d t} = y

Solve separable ODE:

y = \frac{c _{1} 14 t - 15}{14 t - 1}

y = \frac{c _{1} 14 t - 15}{14 t - 1}

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

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

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

x y^{3} d x + e^{- x} d y = 0

\frac{d y}{d x} = x^{2} y, y > 0

What is the solution for (t^2-16t+15)(dy)/(dt)=y ?
The solution for (t^2-16t+15)(dy)/(dt)=y is y=(c_{1}\sqrt[14]{t-15})/(\sqrt[14]{t-1)}