What is the solution for ty^'+2y=t^2-t+1,y(1)=0.5 ?

The solution for ty^'+2y=t^2-t+1,y(1)=0.5 is y=(t^2)/4-t/3+1/2+(0.08333…)/(t^2)

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ty^'+2y=t^2-t+1,y(1)=0.5

t y^{^{'}} + 2 y = t^{2} - t + 1, y (1) = 0.5

y = \frac{t ^{2}}{4} - \frac{t}{3} + \frac{1}{2} + \frac{0.08333 \dots}{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{0.08333 \dots}{t ^{2}}

y = \frac{t ^{2}}{4} - \frac{t}{3} + \frac{1}{2} + \frac{0.08333 \dots}{t ^{2}}

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What is the solution for ty^'+2y=t^2-t+1,y(1)=0.5 ?
The solution for ty^'+2y=t^2-t+1,y(1)=0.5 is y=(t^2)/4-t/3+1/2+(0.08333…)/(t^2)