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}}

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)