Fun With Calculus 8

Consider the solutions are of the form, $y_{1}=e^{r_{1}x} , y_{2}=e^{r_{2}x}$
In general they are of the form $y=e^{rx}$ Differentiate and substitute in the given equation to get, $e^{rx}(r^2-4r+3) =0$ Since $e^{rx}$ is never $0$ then we will have the characteristic equation $r^2-4r+3=0$ whose solutions are $r_{1} =1, r_{2} =3$ .

From the principle of superposition the general solution is of the form, $y=C_{1}e^{3x}+C_{2} e^{x}$ . From the given initial values we have two equations, $3C_{1} +C_{2}=10,C_{1} +C_{2} =6$ Therefore $C_{1}=2, C_{2}=4$
Therefore general solution is, $y=2e^{3x }+4e^{x}$