A linear recurrance is a lot like a linear differerential equation- we can find the general solution to the associated "homogeneous" equation and a solution to the entire equation and add to get the general solution to the entire equation.
Here, the associated homogeneous equation is just T(n)= aT(n-1). If T(0)= C, T(1)= Ca, T(2)= (Ca)a= Ca^2, etc. In other words the general solution to the associated Homogeneous equation is T(n)= Ca^n. Now we need just a single solution to T(n)= aT(n-1)+ bn. If T(0)= 0, T(1)= aT(0)+ b= b, T(2)= aT(1)+ b= ab+ b, T(3)= aT(2)+ b= a^2b+ ab= (a^2+ a)b, T(4)= aT(3)+ b= (a^3+ a^2)b+ b= (a^3+ a^2+ a)b. See the pattern? T(n)= (a^(n-1)+ a^(n-2)+ ...+ a^2+ a)b.
Putting those together T(n)= Ca^n+ b(a^(n-1)+ a^(n-2)+ ...+ a^2+ a)
T(1)= 1 then becomes Ca+ b= 1 so that C= (1-b)/a.