Toronto Math Forum

Find the general solution of the differential equation
$$
16 y^{\prime \prime}+24 y^{\prime}+9 y=0
$$

$$
\begin{array}{l}{16 r^{2}+24 r+9=0} \\ {(4 r+3)^{2}=0} \\ {r_{1}=r_{2}=-\frac{3}{4}} \\ {\therefore y=c_{1} e^{-\frac{3}{4} t}+C_{2} t e^{-\frac{3}{4} t}}\end{array}
$$

Find the general solution of the given differential equation.
$$
2 y^{\prime \prime}-3 y^{\prime}+y=0
$$

$$
\begin{array}{l}{\text { Assume } y=e^{r t}} \\ {\text { then } 2 r^{2}-3 r+1=0} \\ {\qquad(2 r-1)(r-1)=0} \\ {\qquad r_{1}=\frac{1}{2}, r_{2}=1}\end{array}
$$

$\because$ The general solution follows the form
$$
y=c_{1} e^{r_{1} t}+c_{2} e^{r_2 t}
$$
$$
\therefore \quad y=c_{1} e^{\frac{t}{2}}+c_{2} e^{t}
$$