Find general homogeneous solution I have the following exercise:
" Exercise 1) 
Consider the network described by
$$  \frac{L}{R}y'+y=x(t)  $$
a)Find the general homogeneous solution
b) Find a particular solution for the following cases:
i. $x(t) = 1$
ii. $x(t) = t$
iii. $x(t) = 6t+4$
iv. $x(t) = sin(t)$
v. $x(t) = sin(t+5)$"

So I started by calculating the integrating​ factor (which I will symbolize as "u"). I found $u=e^{(R/L)t}$
Then I try to find the solution as follows:
$$ y=\frac{\int{ux(t)dt}}{u} $$
so I get:
$$ y=\frac{\frac{R}{L}\int{x(t)e^{(R/L)t}dt}}{e^{(R/L)t}} $$
Here is where I got stuck. I don't know what to do. I know that the correct solution is $y(t)=Ce^{(-R/L)t}$ but how do I get there from where I am stuck?
 A: When $\alpha$, $\beta$ and $\varphi$ are constants where $\alpha\ne0$:
$$\alpha\cdot\text{y}'\left(t\right)+\beta\cdot\text{y}\left(t\right)=\varphi\cdot\text{x}\left(t\right)\space\Longleftrightarrow\space\text{y}'\left(t\right)+\frac{\beta}{\alpha}\cdot\text{y}\left(t\right)=\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag1$$
Let:
$$\rho\left(t\right)=\exp\left\{\int\frac{\beta}{\alpha}\space\text{d}t\right\}=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\tag2$$
Multuply both sides by $\rho\left(t\right)$:
$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}'\left(t\right)+\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\beta}{\alpha}\cdot\text{y}\left(t\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag3$$
Now, substitute:
$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\beta}{\alpha}=\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\right)\tag4$$
So, we get:
$$\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}'\left(t\right)+\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\right)\cdot\text{y}\left(t\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag5$$
Apply the reverse product rule, to the LHS:
$$\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}\left(t\right)\right)=\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\tag6$$
Now, integrate with respect to $t$:
$$\int\frac{\partial}{\partial t}\left(\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{y}\left(t\right)\right)\space\text{d}t=\int\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\frac{\varphi}{\alpha}\cdot\text{x}\left(t\right)\space\text{d}t\tag7$$
So, we get:
$$\text{y}\left(t\right)=\exp\left\{-\frac{\beta\cdot t}{\alpha}\right\}\cdot\left\{\frac{\varphi}{\alpha}\cdot\int\exp\left\{\frac{\beta\cdot t}{\alpha}\right\}\cdot\text{x}\left(t\right)\space\text{d}t+\text{K}\right\}\tag8$$

Example 1
For the constants, set $\alpha=\frac{\text{L}}{\text{R}}$, $\beta=1$ and $\varphi=1$ and $\text{x}\left(t\right)=1$:
$$\text{y}\left(t\right)=\exp\left\{-\frac{1\cdot t}{\left(\frac{\text{L}}{\text{R}}\right)}\right\}\cdot\left\{\frac{1}{\left(\frac{\text{L}}{\text{R}}\right)}\cdot\int\exp\left\{\frac{1\cdot t}{\left(\frac{\text{L}}{\text{R}}\right)}\right\}\cdot1\space\text{d}t+\text{K}\right\}=$$
$$\exp\left\{-\frac{\text{R}\cdot t}{\text{L}}\right\}\cdot\left\{\frac{\text{R}}{\text{L}}\cdot\int\exp\left\{\frac{\text{R}\cdot t}{\text{L}}\right\}\space\text{d}t+\text{K}\right\}\tag9$$
