First Order Differential Equation: $ty' + ty = 1-y$ $$ty' + ty = 1-y $$
I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help! :( 
I know the first step would be to divide both sides by $t$, which would give:
$$y' + y = (1-y)/t$$
 A: Let's try moving the $y$ terms to the LHS:
$$ty'+y(t+1)=1$$
Then divide by $t$:
$y'+y\left(1+\frac{1}{t}\right)=\frac{1}{t}$
Next we want to find the integrating factor: Hint Below

 $\mu(t)=e^t t$

Using the integrating factor, you can arrive at the general solution: 

 $y=\dfrac{e^{-t}(e^t+c)}{t}$

A: $ty' +(t+1) y = 1\\
y' +\frac{(t+1)}{t} y = \frac 1t$
Integrating factor $= e^{\int \frac{(t+1)}{t} dt}$
$\int \frac{(t+1)}{t} dt = t + ln t\\
e^{t+lnt} = te^t$
We don't need the arbitrary constant for our integrating factor.
$te^ty' +(t+1) e^t y = e^t$
Integrate both sides.  We have chosen the integrating factor such that the right side integrtes nicely.
$t e^t y = e^t+c\\
y = \frac 1t + \frac ct e^{-t}$
A: $$ty'(t)+ty(t)=1-y(t)\Longleftrightarrow$$
$$y'(t)+\frac{y(t)(t+1)}{t}=\frac{1}{t}\Longleftrightarrow$$

Let $r(t)=\exp\left[\int\frac{t+1}{t}\space\text{d}t\right]=te^t$.
Multiply both sides by $r(t)$:

$$\left(te^t\right)y'(t)+y(t)\left(e^t(t+1)\right)=e^t\Longleftrightarrow$$

Substitute $e^t(t+1)=\frac{\text{d}}{\text{d}t}\left(te^t\right)$:

$$\left(te^t\right)y'(t)+\frac{\text{d}}{\text{d}t}\left(te^t\right)y(t)=e^t\Longleftrightarrow$$

Apply the reverse product rule:

$$\frac{\text{d}}{\text{d}t}\left(ty(t)e^t\right)=e^t\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}t}\left(ty(t)e^t\right)\space\text{d}t=\int e^t\space\text{d}t\Longleftrightarrow$$
$$ty(t)e^t=e^t+\text{C}\Longleftrightarrow$$
$$y(t)=\frac{e^t+\text{C}}{te^t}\Longleftrightarrow$$
$$y(t)=\frac{\text{C}e^{-t}}{t}+\frac{1}{t}$$
