4.47 Proof (ii) Double real root

(ii) The auxiliary equation is

$$(s-p)^{2}=s^{2}-2ps+p^{2}=0,$$

since we have a double root. Consider

$$w(x)=e^{-px}y(x)$$

with

$$w^{\prime}(x)=e^{-px}y^{\prime}(x)-pe^{-px}y(x)$$

$$w^{\prime\prime}(x)=e^{-px}y^{\prime\prime}(x)-2pe^{-px}y^{\prime}(x)+p^{2}e^{-px}y(x)$$

where

$$y^{\prime\prime}(x)-2py^{\prime}(x)+p^{2}y(x)=0,$$

so

$$w^{\prime\prime}(x)=0.$$

Then $w(x)=Ax+B$, so

$$y(x)=e^{px}(Ax+B).$$

Then

$$y^{\prime}(x)=pe^{px}(Ax+B)+Ae^{px}$$

so

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