t(dy/dt) + y = 2t?

Question

Find the general solution of the following first-order linear differential equation:

Answer 1

t(dy/dt)+y=0 so -dy/y= dt/t and
-lnIyI = ln I t I +c so lnIt*yI = -c
so Iy*tI=C now find a particular solution of vthe complete equation
It seems to be y = a*t+b
so
a*t+at+b=2t so a= 1 and b=0 and y = t
so the complete solution is
y=C/t+t

Answer 2

Note that the lefthand side is the derivative of the product ty, i.e. d/dt(ty) = t(dy/dt) + y.
So we've got
d/dt(ty) = 2t
and integrating with respect to t gives

ty = t^2 + C
so the general solution is
y = t + C/t.

Note this procedure works for all first-order linear DEs
dy/dt + a(t)y = b(t).
Multiplying through by e^(integral a(t) dt) makes the lefthand side the derivative of a product

d/dt(e^(integral a(t) dt)*y) = b(t)*e^(integral a(t) dt)
and then it's a matter of whether we can do this integration!