Question

Let $Î\pm (t)$ and $β(t)$ be differentiable functions on R such that $Î\pm (0)=2$ and $β(0)=1$. If $Î\pm (t)+{β}^{′}(t)=1$ and ${Î\pm }^{′}(t)+β(t)=1$ for all $t∈[0,∞)$, then the value of $Î\pm (\ln 2)$ is expressed in the lowest form as $\frac{p}{q}$. Find the value of $(p−q)$.

Answer 1

Answer: 5

$Î\pm (t)+{β}^{′}(t)=1$....(1)
${Î\pm }^{′}(t)+β(t)=1$....(2)
Add (1) and (2)
$(Î\pm +β)+\left({Î\pm }^{′}+{β}^{′}\right)=2$
$∴(Î\pm +β)+(Î\pm +β{)}^{′}=2$
$(Î\pm +β)=y$
$∴\frac{dy}{dt}+y=2\text{ I.F. }=e^t$
$yâ‹\dots e^t=2e^t+C$
$(Î\pm (t)+β(t))e^t=2e^t+C$
$\text{ Put }t=0$
$2+1=2+C$
$C=1$
$∴Î\pm (t)+β(t)=2+e^{−t}$....(3)
|||ly $(1)−(2)$
$(Î\pm −β)−(Î\pm −β{)}^{′}=0$
$(Î\pm −β{)}^{′}−(Î\pm −β)=0$
$\frac{dy}{dt}−y=0\text{ I.F. }=e^{−t}$
$y{e}^{−t}=C$
$y=C{e}^t$
$Î\pm (t)−β(t)=C{e}^t\text{ Put }t=0$
$2−1=C∴C=1\$<br/>$\alpha( t )-\beta( t )= e ^{ t } $.....(4)<br/>$\therefore $Put$t =\ln 2$in(3)\&(4)andadd<br/>$2 \alpha(\ln 2)=2+\frac{1}{2}+2=4+\frac{1}{2}=\frac{9}{2} $<br/>$\alpha(\ln 2)=\frac{9}{4}=\frac{p}{q} $<br/>$\therefore p-q=9-4=5$