Answer:
Step-by-step explanation:
Given is a differential equation as
[tex]ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0[/tex]
Divide this by t to get in linear form
[tex]y'+6y/t = t-1+1/t[/tex]
This is of the form
y' +p(t) y = Q(t)
where p(t) = 1/t[tex]e^(\int 1/tdt) = t[/tex]
So solution would be
[tex]yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C[/tex]
siubstitute y(1) = 16
[tex]16 = 16^3/3-128+1+C\\C = -1206[/tex]
