Question

Let $\alpha(t)$ and $\beta(t)$ be differentiable functions on R such that $\alpha(0)=2$ and $\beta(0)=1$. If $\alpha(t)+\beta^{\prime}(t)=1$ and $\alpha^{\prime}(t)+\beta(t)=1$ for all $t \in[0, \infty)$, then the value of $\alpha(\ln 2)$ is expressed in the lowest form as $\frac{p}{q}$. Find the value of $(p-q)$.

Answer 1

$\alpha(t)+\beta^{\prime}(t)=1$....(1)
$\alpha^{\prime}( t )+\beta( t )=1$....(2)
Add (1) and (2)
$(\alpha+\beta)+\left(\alpha^{\prime}+\beta^{\prime}\right)=2 $
$\therefore (\alpha+\beta)+(\alpha+\beta)^{\prime}=2 $
$(\alpha+\beta)=y$
$\therefore \frac{ dy }{ dt }+ y =2 \text { I.F. }= e ^{ t } $
$y \cdot e ^{ t }=2 e ^{ t }+ C$
$ (\alpha( t )+\beta( t )) e ^{ t }=2 e ^{ t }+ C $
$ \text { Put } t =0 $
$2+1=2+ C$
$ C =1$
$\therefore \alpha( t )+\beta( t )=2+ e ^{- t }$....(3)
|||ly $(1) - (2)$
$(\alpha-\beta)-(\alpha-\beta)^{\prime}=0 $
$(\alpha-\beta)^{\prime}-(\alpha-\beta)=0 $
$\frac{ dy }{ dt }- y =0 \text { I.F. }= e ^{- t } $
$ye ^{- t }= C $
$y=C e^t $
$\alpha( t )-\beta( t )= C e ^{ t } \text { Put } t =0 $
$2-1= C \therefore C =1 \$
$\alpha( t )-\beta( t )= e ^{ t } $.....(4)
$\therefore $ Put $t =\ln 2$ in (3) \& (4) and add
$2 \alpha(\ln 2)=2+\frac{1}{2}+2=4+\frac{1}{2}=\frac{9}{2} $
$\alpha(\ln 2)=\frac{9}{4}=\frac{p}{q} $
$\therefore p-q=9-4=5$