First Order differential equation ... How did they do it ? I was studying "continuous and discrete signals and systems" by Samir S. Soliman where I encountered with this first order differential equation:
$$
\frac{dy(t)}{dt} + \frac{R_1R_2}{L(R_1+R_2)}y(t) = \frac{R_2}{L(R_1+R_2)}x(t)
$$
 They mention that in next line " To compute an explicit expression for y(t) in terms of x(t), we must solve the differential equation for an arbitrary input x(t) applied for $\ t >= t_o $ . The complete solution is of the form "
$$
y(t) = y(t_o) exp[-\frac{R_1R_2}{L(R_1+R_2)}(t-t_o)] + \frac{R_2}{L(R_1+R_2)}\int_{t_o}^{t}exp[-\frac{R_1R_2}{L(R_1+R_2)}(t-\tau)]x(\tau) d\tau ;t >= t_o
$$
But there is no explanation how they get this result in detail. How did the solve this? How did they get this limit or$\ \tau $ . 
Please explain this total solution in detail (step by step  approach to solve this kind of problems is appreciated ) .
I have introductory knowledge on differential equation solving but never encountered this kind of problem.
 A: General Solution (using the method of integrating factor):
Let $\alpha = \frac{R_1R_2}{L(R_1+R_2)}$ and $\beta=\frac{R_2}{L(R_1+R_2)}$. The linear inhomogeneous differential equation becomes:
$$\frac{dy(t)}{dt}+\alpha y(t) = \beta x(t)$$
Let $\mu(t)  =  e^{\int \alpha dt}  =  e^{\alpha t}$ and multiply both sides by $\mu(t)$:
$$e^{\alpha t} \frac{dy(t)}{dt}+\alpha e^{\alpha t} y(t)  =  \beta e^{\alpha t} x(t)$$
Substitute $\alpha e^{\alpha t}  =  \frac{d}{dt}(e^{\alpha t})$:
$$e^{\alpha t} \frac{dy(t)}{dt}+\frac{d}{dt}(e^{\alpha t}) y(t)  =  \beta e^{\alpha t} x(t)$$
Apply the reverse product rule $g \frac{df}{dt}+f \frac{dg}{dt}  =  \frac{d}{dt}(f g)$ to the left-hand side:
$$\frac{d}{dt}(e^{\alpha t} y(t))  =  \beta e^{\alpha t} x(t)$$
Integrate both sides with respect to $t$:
$$\int \frac{d}{dt}(e^{\alpha t} y(t)) dt  =  \int \beta e^{\alpha t} x(t) dt$$
Evaluate the integrals:
$$e^{\alpha t} y(t)  =  \beta  \int e^{\alpha t} x(t) dt+C$$ where $C$ is an arbitrary constant.
Divide both sides by $\mu(t)  =  e^{\alpha t}$ to get the general solution:
$$y(t)  =  \beta e^{-\alpha t}  \int e^{\alpha t} x(t) dt+C e^{-\alpha t}$$

I will leave the rest for you. Give it a shot.
