solve $y'=ay+b$ I have this differential equation which I want to solve
$\displaystyle\frac{dT_i}{dt}=\frac{1}{RC}(T_a-T_i)+\frac{1}{C} \Phi_h$
I know it is in the form 
$y'=ay+b$
But how can I solve it ? 
 A: If you have $$y'(x)=ay(x)+b(x),$$for constant $a$, then the simpliest way is to multiply the equation by $e^{-ax}$ (nonzero everywhere) and deduce that 
$$(y(x)e^{-ax})' = e^{-ax}b(x),$$which implies that
$$y(x)e^{-ax} - y(0) = \int_0^x e^{-as}b(s)ds.$$
Finally,
$$y(x) = y(0)e^{ax} + \int_0^x e^{-a(x-s)}b(s)ds.$$
A: As a separable equation:
\begin{align}
{dy\over dx}&=ay+b\\
{dy\over ay+b}&=dx\\
\int {dy\over ay+b}&=\int\,dx\\
{1\over a}\ln|ay+b|&=x+C\\
\ln|ay+b|&=ax+aC\\
e^{\ln|ay+b|}&=e^{ax+aC}\\
|ay+b|&=e^{ax}e^{aC}\\
ay+b&=\pm e^{aC}e^{ax}\\
ay+b&=De^{ax}, \quad D\not= 0\\
y&=Ke^{ax}-{b\over a}
\end{align}
You can also solve it via an integrating factor:
\begin{align}
y'-ay&=b \qquad\quad\therefore\text{integrating factor is }e^{\int -a\,dx}=e^{-ax}\\
e^{-ax}(y'-ay)&=e^{-ax}b\\
[e^{-ax}y]'&=be^{-ax}\\
\int[e^{-ax}y]'\,dx&=\int be^{-ax}\,dx\\
e^{-ax}y&={be^{-ax}\over -a}+C\\
y&=-{b\over a}+Ce^{ax}
\end{align}
A: $$\dfrac{1}{a}\int \dfrac{dy}{y+\dfrac{b}{a}}=\dfrac{1}{a}\int\dfrac{d\left(y+\dfrac{b}{a}\right)}{y+\dfrac{b}{a}}=\int dx$$
