What is the solution for ty^'+3y=t^2-t+7 ?

The solution for ty^'+3y=t^2-t+7 is y=(t^2)/5-t/4+7/3+(c_{1})/(t^3)

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ty^'+3y=t^2-t+7

t y^{^{'}} + 3 y = t^{2} - t + 7

y = \frac{t ^{2}}{5} - \frac{t}{4} + \frac{7}{3} + \frac{c _{1}}{t ^{3}}

Solution steps

t y^{'} (t) + 3 y = t^{2} - t + 7

Solve linear ODE:

y = \frac{t ^{2}}{5} - \frac{t}{4} + \frac{7}{3} + \frac{c _{1}}{t ^{3}}

y = \frac{t ^{2}}{5} - \frac{t}{4} + \frac{7}{3} + \frac{c _{1}}{t ^{3}}

x \to 0 lim (\frac{1}{x})

x \to 2 lim (x^{3} - 4 x)

f (x) = x^{2} - 5 x

x \to 0 lim (\frac{1}{x} + 3)

\frac{d}{d x} (e^{(x + y)})

What is the solution for ty^'+3y=t^2-t+7 ?
The solution for ty^'+3y=t^2-t+7 is y=(t^2)/5-t/4+7/3+(c_{1})/(t^3)