What is the solution for (t^2-20t+51)(dy)/(dt)=y ?

The solution for (t^2-20t+51)(dy)/(dt)=y is y=(c_{1}\sqrt[14]{t-17})/(\sqrt[14]{t-3)}

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(t^2-20t+51)(dy)/(dt)=y

(t^{2} - 20 t + 51) \frac{d y}{d t} = y

Solution steps

(t^{2} - 20 t + 51) \frac{d y}{d t} = y

Solve separable ODE:

\frac{1}{2 x} + \frac{1}{2 y} \frac{d y}{d x} = 0

e^{2 t} \frac{d x}{d t} = cos^{2} (x)

x \frac{d y}{d x} - y = 6 x^{9} y

u^{^{'}} = e^{6 H (t) + 9 t}

4 y^{^{'}} + 2 y = 120, y (0) = 1

What is the solution for (t^2-20t+51)(dy)/(dt)=y ?
The solution for (t^2-20t+51)(dy)/(dt)=y is y=(c_{1}\sqrt[14]{t-17})/(\sqrt[14]{t-3)}