Question

Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0

Answer 1

Answer:

Step-by-step explanation:

Given is a differential equation as

$ty' + 6y = t^2 - t + 1, y(1) = 1 6 , t > 0$

Divide this by t to get in linear form

$y'+6y/t = t-1+1/t$

This is of the form

y' +p(t) y = Q(t)

where p(t) = 1/t$e^(\int 1/tdt) = t$

So solution would be

$yt = \int t^2-t+1 dt\\= t^3/3-t^2/2+t+C$

siubstitute y(1) = 16

$16 = 16^3/3-128+1+C\\C = -1206$