What is the solution for ty^'+4y=t^2-t+7 ?

The solution for ty^'+4y=t^2-t+7 is y=(t^2)/6-t/5+7/4+(c_{1})/(t^4)

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ty^'+4y=t^2-t+7

t y^{^{'}} + 4 y = t^{2} - t + 7

y = \frac{t ^{2}}{6} - \frac{t}{5} + \frac{7}{4} + \frac{c _{1}}{t ^{4}}

Solution steps

t y^{'} (t) + 4 y = t^{2} - t + 7

Solve linear ODE:

y = \frac{t ^{2}}{6} - \frac{t}{5} + \frac{7}{4} + \frac{c _{1}}{t ^{4}}

y = \frac{t ^{2}}{6} - \frac{t}{5} + \frac{7}{4} + \frac{c _{1}}{t ^{4}}

t^{2} sin^{2} (t)

\frac{d y}{d x} = \frac{x}{25 y}, y (0) = - 5

2.5 (1 + e^{- 0.1 t})

\frac{d ^{2}}{d x ^{2}} (sin (3 x))

\int 3 sin^{3} (x co) s^{7} x d x

What is the solution for ty^'+4y=t^2-t+7 ?
The solution for ty^'+4y=t^2-t+7 is y=(t^2)/6-t/5+7/4+(c_{1})/(t^4)