What is the solution for ty^'+2y=t^2-t+1,y(1)= 1/2 ?

The solution for ty^'+2y=t^2-t+1,y(1)= 1/2 is y=(t^2)/4-t/3+1/2+1/(12t^2)

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ty^'+2y=t^2-t+1,y(1)= 1/2

t y^{^{'}} + 2 y = t^{2} - t + 1, y (1) = \frac{1}{2}

y = \frac{t ^{2}}{4} - \frac{t}{3} + \frac{1}{2} + \frac{1}{12 t ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{t ^{2}}{4} - \frac{t}{3} + \frac{1}{2} + \frac{1}{12 t ^{2}}

y = \frac{t ^{2}}{4} - \frac{t}{3} + \frac{1}{2} + \frac{1}{12 t ^{2}}

f (x) = (x^{3} + 2) (3 x^{2} - 4 x + 2), (1, 3)

\int \frac{3 x ^{2} + 3 x + 1}{x ^{3} + 2 x ^{2} + 2 x + 1} d x

(1, 2), (5, - 3)

y = \frac{1}{x + 1}

\int \frac{7 x ^{2}}{x ^{2} - 4} d x

What is the solution for ty^'+2y=t^2-t+1,y(1)= 1/2 ?
The solution for ty^'+2y=t^2-t+1,y(1)= 1/2 is y=(t^2)/4-t/3+1/2+1/(12t^2)