What is the solution for ty^'+4y=t^2-t+5 ?

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

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ty^'+4y=t^2-t+5

t y^{^{'}} + 4 y = t^{2} - t + 5

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

Solution steps

t y^{'} (t) + 4 y = t^{2} - t + 5

Solve linear ODE:

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

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

\int_{1}^{8} (3 x) d x

\frac{\partial}{\partial t} (1 + 2 t)

\int (x^{2} + x + 1) (e^{x}) d x

x y - 3 y^{2} = 2, (7, 2)

x \to 0 lim (3 x - 5)

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