Solving recurrence relation of the form $T(n)=T\left(\frac{nb}{a}\right)+T\left(\frac{(n-b)c}{a}\right)+n$ How do we solve a recurrence of the form:
$T(n)=T\biggl(\dfrac{nb}{a}\biggr)+T\biggl(\dfrac{(n-b)c}{a}\biggr)+n$ ?
I tried substituting $q=\dfrac{a}{b},\ r=\dfrac{a}{c}, \ s=\dfrac{bc}{a}$ to obtain:
$T(n)=T\left(\dfrac{n}{q}\right)+T\left(\dfrac{n}{r}-s\right)+n$ 
but could not proceed further. 
This question solves a somewhat similar problem but it has a convenient form which makes for a neat substitution.
 A: Assume $a,b,c\neq0$ and $\dfrac{b}{a}\neq1$ to maintain the key meaning of this question.
$T(n)=T\biggl(\dfrac{nb}{a}\biggr)+T\biggl(\dfrac{(n-b)c}{a}\biggr)+n$
$T(n)-T\biggl(\dfrac{nb}{a}\biggr)-T\biggl(\dfrac{(n-b)c}{a}\biggr)=n$
Getting the particular solution part is very easy when $a-b-c\neq0$ .
Let $T_p(n)=An+B$ ,
Then $An+B-\biggl(\dfrac{Anb}{a}+B\biggr)-\biggl(\dfrac{A(n-b)c}{a}+B\biggr)\equiv n$
$\dfrac{(a-b-c)An}{a}+\dfrac{bcA}{a}-B\equiv n$
$\therefore\begin{cases}\dfrac{(a-b-c)A}{a}=1\\\dfrac{bcA}{a}-B=0\end{cases}$
$\begin{cases}A=\dfrac{a}{a-b-c}\\B=\dfrac{bc}{a-b-c}\end{cases}$
$\therefore T_p(n)=\dfrac{an+bc}{a-b-c}$
But getting the particular solution part is not easy when $a-b-c=0$ .
For getting the complementary solution part, we should handle the equation $T_c(n)-T_c\biggl(\dfrac{nb}{a}\biggr)-T_c\biggl(\dfrac{(n-b)c}{a}\biggr)=0$
Even let $T_c(n)=\int_{-\infty}^\infty e^{nt}K(t)~dt$ ,
Then $\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty e^{\frac{nbt}{a}}K(t)~dt-\int_{-\infty}^\infty e^{\frac{(n-b)ct}{a}}K(t)~dt=0$
$\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty e^{\frac{bnt}{a}}K(t)~dt-\int_{-\infty}^\infty e^{\frac{cnt}{a}}e^{-\frac{bct}{a}}K(t)~dt=0$
$\int_{-\infty}^\infty e^{nt}K(t)~dt-\text{sgn}\left(\dfrac{a}{b}\right)\int_{-\infty}^\infty e^{nt}K\left(\dfrac{at}{b}\right)d\left(\dfrac{at}{b}\right)-\text{sgn}\left(\dfrac{a}{c}\right)\int_{-\infty}^\infty e^{nt}e^{-bt}K\left(\dfrac{at}{c}\right)d\left(\dfrac{at}{c}\right)=0$
$\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)e^{nt}K\left(\dfrac{at}{b}\right)dt-\int_{-\infty}^\infty\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{nt}e^{-bt}K\left(\dfrac{at}{c}\right)dt=0$
$\int_{-\infty}^\infty\left(K(t)-\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)K\left(\dfrac{at}{b}\right)-\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{-bt}K\left(\dfrac{at}{c}\right)\right)e^{nt}~dt=0$
$\therefore K(t)-\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)K\left(\dfrac{at}{b}\right)-\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{-bt}K\left(\dfrac{at}{c}\right)=0$
But still very difficult to solve in general.
