Talk:Residential module user's guide

ETP solution for unusual cases
--Dchassin 03:08, 18 March 2013 (UTC)

It should be noted that the solution of ETP in Equation (4) in only valid when the characteristic equation (3) has two real roots. However, if the characteristic equation has a single repeated root $$r = r_1 = r_2 \!$$, then the solution is of the form


 * $$T_A = A_1 e^{r t} + A_2 t e^{r t} \!$$.

If the characteristic equation has complex roots $$r_{1,2} = \lambda \plusmn i \mu$$ then the solution is


 * $$T_A = A_1 e^{\lambda t} cos(\mu t) + A_2 e^{\lambda t} sin(\mu t) \!$$.