Answer
Q: $ty'+(t+1)y=t$, $y(ln2)=1$, $t>0$, solve the initial value problem
Answer: $y=\frac{(t-1+2e^{-t})}{t}$
Work Step by Step
Because this is a differential equations course question, I have omitted most of the explanation of the calculus, if do not remember how to do the calculus then keep a formula sheet of common calculus operations handy. This is of the form $p(t)y'+g(t)y=h(t)$, the first thing we need to do is to find a $p(t)$ such that $d(p(t)y)=p(t)y'+g(t)y$.
step 1, divide entire equation by t
$y'+\frac{t+1}{t}y$=1
step 2, multiply entire equation by integrating factor $µ(t)$
$µ(t)y'+µ(t)\frac{t+1}{t}y=µ(t)$
step 3, in order to solve this equation, we must make sure that the following is true: $µ(t)y'+µ(t)\frac{t+1}{t}y=d(µ(t)y)$, thus, $µ(t)\frac{t+1}{t}=d(µ(t))$ we must do this to isolate $y$
step 4, solve for µ(t), use the fact that $d(ln(x))=\frac{x'}{x}$
$\frac{t+1}{t}=\frac{d(µ(t))}{µ(t)}$
$\int\frac{t+1}{t}dt=\int{(1+t^{-1})}dt=\int\frac{d(µ(t))}{µ(t)}dt$
$t+ln(t)+C=ln(µ(t))$, now raise everything as an exponent of $e$
$e^{t+ln(t)+C}=e^{ln(µ(t))}$, using exponent properties we get
$e^{t}e^{ln(t)}e^{C}=µ(t)$, e^C is still an arbitrary constant called k (the name change is insignificant but my reason for doing so will become clear in the next step)
$kte^{t}=µ(t)$
Step 5, $d(µ(t)y)=µ(t)$
therefore, $µ(t)y+C=\intµ(t)dt$ or $y=\frac{\intµ(t)dt+C}{µ(t)}$
which is $y=\frac{\int{kte^{t}}dt+C}{kte^{t}}$, and thus(note, C/k is still C because C is arbitrary)
$y=\frac{\int{te^{t}}dt+C}{te^{t}}$, this is our explicit solution, but we must evaluate the integral before we can plug in the initial value
$y=\frac{te^{t}-e^t+C}{te^{t}}$, this is our fully evaluated explicit solution.
Step 6, $y(ln(2))=1$, so $1=\frac{(ln(2))e^{ln2}-e^{ln2}+C}{(ln(2))e^{ln2}}$
$1=\frac{2ln(2)-2+C}{2ln(2)}$,
$1=1+\frac{-2+C}{2ln(2)}$
$0=\frac{-2+C}{2ln(2)}$
$0=-2+C$
$C=2$, therefore
$y=\frac{te^{t}-e^t+2}{te^{t}}$ or $y=\frac{t-1+2e^{-t}}{t}$
