Originally Posted by
HallsofIvy Are you taking a course in differential equations? This is a very basic, simple example. Its "characteristic equation" is $\displaystyle 12r^2- 3= 0$ which has roots $\displaystyle r= \pm\frac{1}{2}$ so the "general solution" to the differential equation is $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$ (or equivalently, $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$).
From the formula $\displaystyle y(t)= Ae^{t/2}+ Be^{-t/2}$, $\displaystyle y'(t)= (A/2)e^{t/2}- (B/2)e^{-t/2}$ so y(0)= A+ B= 3 and y'(0)= A/2- B/2= 1/2.
From the formula $\displaystyle y(t)= C cosh(t/2)+ D sinh(t/2)$, $\displaystyle y'(t)= (C/2)sinh(t/2)+ (D/2)cosh(t/2)$ so y(0)= C= 3 and y'(0)= D/2= 1/2.