What is the solution for y=(t^2-20t+84)y^' ?

The solution for y=(t^2-20t+84)y^' is y=(c_{1}\sqrt[8]{t-14})/(\sqrt[8]{t-6)}

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y=(t^2-20t+84)y^'

y = (t^{2} - 20 t + 84) y^{^{'}}

y = \frac{c _{1} 8 t - 14}{8 t - 6}

Solution steps

y = (t^{2} - 20 t + 84) y^{'} (t)

Solve separable ODE:

y = \frac{c _{1} 8 t - 14}{8 t - 6}

y = \frac{c _{1} 8 t - 14}{8 t - 6}

2 x \cdot sin (2 y) + 2 x^{2} \cdot cos (2 y) y^{^{'}} = 0

y^{^{'}} = \frac{2 x}{- 8 y - 10}

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

\frac{d y}{d x} = 160 + 0.02 \cdot y, y (0) = 2000

\frac{d y}{d x} y = cos (tan (7 x + 3))

What is the solution for y=(t^2-20t+84)y^' ?
The solution for y=(t^2-20t+84)y^' is y=(c_{1}\sqrt[8]{t-14})/(\sqrt[8]{t-6)}