Question

determine the values of r for which the given differential equation has solutions of the form y = e^rt

10.y′′\:+\:y′\:−\:6y\:=\:0

Answer 1

Let us determine the values of $r$ for which the differential equation $y''\:+\:y'\:−\:6y\:=\:0$ has solutions of the form $y = e^{rt}.$ Taking into account that the characteristic equation $r^2+r-6=0$ is equivalent to $(r-2)(r+3)=0,$ we conclude that it has the roots $r_1=2$ and $r_2=-3.$ Therefore, for the values $r_1=2$ and $r_2=-3$ the differential equation $y''\:+\:y'\:−\:6y\:=\:0$ has solutions of the form $y = e^{r_1t}$ and $y = e^{r_2t}.$