4.46 Proof (i) Distinct real roots

We have

$$ap^{2}+bp+c=0,$$

so $y_{1}=e^{px}$ is a solution since

$$ay_{1}^{\prime\prime}+by_{1}^{\prime}+cy_{1}=(ap^{2}+bp+c)e^{px}=0;$$

likewise

$$aq^{2}+bq+c=0,$$

so $y_{2}=e^{qx}$ is a solution; then by linear superposition

$$y=Ae^{px}+Be^{qx}$$

is a solution. Also,

$$y^{\prime}(x)=Ape^{px}+Bqe^{qx}$$

so

$$y(0)=A+B,\quad y^{\prime}(0)=Ap+Bq.$$