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

Differential Equation Solution Explained

The question asks us to find the solution $Q(t)$ for the given differential equation:
$$\frac{dQ}{dt}+Q=1$$
with the initial condition $Q=0$ at $t=0$. This is a first-order linear ordinary differential equation.

Identifying the Type of Differential Equation

The given differential equation $\frac{dQ}{dt}+Q=1$ is in the standard form of a first-order linear differential equation, which is generally written as:
$$\frac{dy}{dx}+P(x)y=R(x)$$
In our case, comparing $\frac{dQ}{dt}+Q=1$ with the standard form, we have:

- $y$ replaced by $Q$
- $x$ replaced by $t$
- $P(t)=1$ (the coefficient of $Q$)
- $R(t)=1$ (the function on the right-hand side)

Solving Using the Integrating Factor Method

The integrating factor method is suitable for solving first-order linear differential equations.

1. Calculate the Integrating Factor (IF): The integrating factor is given by the formula: $$IF=e^{\int P(t)dt}$$ Substituting $P(t)=1$: $$IF=e^{\int 1dt}=e^t$$
2. Multiply the Differential Equation by the Integrating Factor: Multiply both sides of the original equation $\frac{dQ}{dt}+Q=1$ by the integrating factor $e^t$: $$e^t\frac{dQ}{dt}+e^{t}Q=e^t\cdot 1$$ $$e^t\frac{dQ}{dt}+Q{e}^t=e^t$$ The left-hand side of this equation is the derivative of the product $(Q\cdot IF)$ with respect to $t$. This is a key property of the integrating factor method: $$\frac{d}{dt}(Q\cdot e^t)=e^t$$
3. Integrate Both Sides: Now, integrate both sides of the equation with respect to $t$: $$\int \frac{d}{dt}(Q{e}^t)dt=\int e^{t}dt$$ $$Q{e}^t=e^t+C$$ where $C$ is the constant of integration.
4. **Solve for $Q(t)$:** To find $Q(t)$, divide both sides by $e^t$: $$Q(t)=\frac{e^t+C}{e^t}$$ $$Q(t)=\frac{e^t}{e^t}+\frac{C}{e^t}$$ $$Q(t)=1+C{e}^{-t}$$
5. Apply the Initial Condition: We are given the initial condition $Q=0$ at $t=0$. Substitute these values into the general solution $Q(t)=1+C{e}^{-t}$: $$0=1+C{e}^{-0}$$ $$0=1+C\cdot 1$$ $$0=1+C$$ Solving for $C$: $$C=-1$$
6. **Write the Final Solution $Q(t)$:** Substitute the value of $C=-1$ back into the general solution $Q(t)=1+C{e}^{-t}$: $$Q(t)=1+(-1)e^{-t}$$ $$Q(t)=1-e^{-t}$$

Conclusion

The solution to the differential equation $\frac{dQ}{dt}+Q=1$ with the initial condition $Q=0$ at $t=0$ is $Q(t)=1-e^{-t}$.