To solve the differential equation [tex]\(\frac{dA}{dt} = \frac{1}{At}\)[/tex] with the initial condition [tex]\(A(1) = 8\)[/tex], follow these steps:
### Step 1: Separate the Variables
The given differential equation is:
[tex]\[
\frac{dA}{dt} = \frac{1}{A t}
\][/tex]
First, we need to separate the variables [tex]\(A\)[/tex] and [tex]\(t\)[/tex]. Multiply both sides by [tex]\(A\)[/tex] and [tex]\(dt\)[/tex]:
[tex]\[
A \, dA = \frac{1}{t} \, dt
\][/tex]
### Step 2: Integrate Both Sides
Integrate both sides to find [tex]\(A\)[/tex] as a function of [tex]\(t\)[/tex]:
[tex]\[
\int A \, dA = \int \frac{1}{t} \, dt
\][/tex]
The integrals are:
[tex]\[
\int A \, dA = \frac{A^2}{2} + C_1 \quad \text{and} \quad \int \frac{1}{t} \, dt = \ln |t| + C_2
\][/tex]
Combining, we get:
[tex]\[
\frac{A^2}{2} = \ln |t| + C
\][/tex]
where [tex]\(C = C_2 - C_1\)[/tex] is a constant of integration.
### Step 3: Solve for [tex]\(A\)[/tex]
Multiply both sides by 2 to isolate [tex]\(A^2\)[/tex]:
[tex]\[
A^2 = 2\ln |t| + 2C
\][/tex]
Let [tex]\(2C\)[/tex] be a new constant, [tex]\(C'\)[/tex]:
[tex]\[
A^2 = 2\ln |t| + C'
\][/tex]
### Step 4: Apply the Initial Condition
Use the initial condition [tex]\(A(1) = 8\)[/tex] to determine the constant [tex]\(C'\)[/tex]:
[tex]\[
8^2 = 2\ln |1| + C'
\][/tex]
[tex]\[
64 = 2(0) + C'
\][/tex]
[tex]\[
C' = 64
\][/tex]
So the equation becomes:
[tex]\[
A^2 = 2\ln |t| + 64
\][/tex]
### Step 5: Solve for [tex]\(A\)[/tex]
Finally, solve for [tex]\(A\)[/tex] by taking the square root of both sides:
[tex]\[
A = \sqrt{2\ln |t| + 64}
\][/tex]
Since [tex]\(A > 0\)[/tex] and [tex]\(\ln |t|\)[/tex] is defined for [tex]\(t > 0\)[/tex], we get the final solution as:
[tex]\[
A(t) = \sqrt{2\ln t + 64}
\][/tex]
Thus, the solution to the given differential equation satisfying the initial condition [tex]\(A(1) = 8\)[/tex] is:
[tex]\[
A(t) = \sqrt{2\ln t + 64}
\][/tex]
