What is the solution for ty^'+8y=t^2-t+1 ?

The solution for ty^'+8y=t^2-t+1 is y=(t^2)/(10)-t/9+1/8+(c_{1})/(t^8)

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ty^'+8y=t^2-t+1

t y^{^{'}} + 8 y = t^{2} - t + 1

y = \frac{t ^{2}}{10} - \frac{t}{9} + \frac{1}{8} + \frac{c _{1}}{t ^{8}}

Solution steps

t y^{'} (t) + 8 y = t^{2} - t + 1

Solve linear ODE:

y = \frac{t ^{2}}{10} - \frac{t}{9} + \frac{1}{8} + \frac{c _{1}}{t ^{8}}

y = \frac{t ^{2}}{10} - \frac{t}{9} + \frac{1}{8} + \frac{c _{1}}{t ^{8}}

\frac{d}{d x} (\frac{1}{( 25 - x ^{2} )})

\int e^{7} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} = 8 x + y, y (0) = 1

\frac{d y}{d x} = \frac{2 x y - y ^{2}}{x ^{2}}

What is the solution for ty^'+8y=t^2-t+1 ?
The solution for ty^'+8y=t^2-t+1 is y=(t^2)/(10)-t/9+1/8+(c_{1})/(t^8)