The function Q(t) satisfies the following ordinary differential equation: dQ/dt = -11Q. What is the general solution of this equation?

The given ordinary differential equation is:

\[
\frac{dQ}{dt} = -11Q
\]

This is a first-order linear differential equation, and it can be solved by separating the variables. Here's the step-by-step solution:

### Step 1: Separate the variables
We want to rearrange the equation so that all terms involving \(Q\) are on one side, and all terms involving \(t\) are on the other side:

\[
\frac{1}{Q} \, dQ = -11 \, dt
\]

### Step 2: Integrate both sides
Now integrate both sides:

\[
\int \frac{1}{Q} \, dQ = \int -11 \, dt
\]

The integrals are:

\[
\ln|Q| = -11t + C
\]

where \(C\) is the constant of integration.

### Step 3: Solve for \(Q\)
To solve for \(Q\), we exponentiate both sides:

\[
|Q| = e^{(-11t + C)} = e^{-11t} e^C
\]

Since \(e^C\) is just a constant, we can rewrite it as a new constant \(A = e^C\):

\[
|Q| = A e^{-11t}
\]

Thus, the general solution is:

\[
Q(t) = A e^{-11t}
\]

where \(A\) is a constant determined by initial conditions.

---

Would you like further details on the steps, or do you have any questions?

### Related Questions:
1. What are the initial conditions needed to determine the constant \(A\)?
2. How does the solution behave as \(t\) increases or decreases?
3. What is the physical interpretation of the constant \(A\) in an applied context?
4. How would the solution change if the coefficient \(-11\) were replaced by a different value?
5. Can you solve similar differential equations with non-constant coefficients?

### Tip: 
For first-order linear differential equations, separation of variables is a powerful and simple technique when the equation is separable.

Discover the general solution to the first-order linear differential equation dQ/dt = -11Q. Using separation of variables, learn how to solve this equation, leading to the exponential solution Q(t) = A e^{-11t}. Suitable for undergraduate calculus students.

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