Question

The solution of the equation $\frac{dQ}{dt}+Q=1$ with $Q=0$ at $t=0$ is

• A. $Q\left(t\right)=e^{-t}-1$
• B. $Q\left(t\right)=1+e^{-t}$
• C. $Q\left(t\right)=1-e^t$
• D. $Q\left(t\right)=1-e^{-t}$

Answer 1

The correct option is D $Q\left(t\right)=1-e^{-t}$

$\frac{dQ}{dt}+Q=1$
$I.F=e^{\int pdt}=e^{\int 1dt}=e^t$
$Q\left(t\right)e^t=\int e^{t}dt+C$
$Q\left(t\right)e^t=e^t+C$
At $t=0,Q=0$
$0=1+C$
$C=-1$
$Q\left(t\right)=1-e^{-t}$