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We are given the differential equation for the current in an electric circuit:

\[
Ri + L \frac{di}{dt} = 0
\]

where \(R\) and \(L\) are constants. We need to show that the solution to this equation is:

\[
i = I e^{-\frac{R}{L}t}
\]

where \( i = I \) when \( t = 0 \).

### Step-by-step Solution:

1. **Rearrange the equation**:
   Starting from the differential equation:
   \[
   Ri + L \frac{di}{dt} = 0
   \]
   Divide through by \(L\):
   \[
   \frac{R}{L} i + \frac{di}{dt} = 0
   \]
   This simplifies to:
   \[
   \frac{di}{dt} = -\frac{R}{L} i
   \]

2. **Separate the variables**:
   To solve the differential equation, we separate the variables \(i\) and \(t\):
   \[
   \frac{1}{i} \, di = -\frac{R}{L} \, dt
   \]

3. **Integrate both sides**:
   Integrating both sides:
   \[
   \int \frac{1}{i} \, di = \int -\frac{R}{L} \, dt
   \]
   The left side gives \(\ln |i|\), and the right side gives \(-\frac{R}{L} t\):
   \[
   \ln |i| = -\frac{R}{L} t + C
   \]
   where \(C\) is the constant of integration.

4. **Solve for \(i\)**:
   Exponentiate both sides to eliminate the logarithm:
   \[
   i = e^{-\frac{R}{L} t + C}
   \]
   This simplifies to:
   \[
   i = e^C e^{-\frac{R}{L} t}
   \]
   Let \( e^C = I \) (where \(I\) is a constant). Therefore:
   \[
   i = I e^{-\frac{R}{L} t}
   \]

5. **Apply the initial condition**:
   We're given that \(i = I\) when \(t = 0\). Substitute \(t = 0\) into the equation:
   \[
   i(0) = I e^{-\frac{R}{L}(0)} = I
   \]
   Hence, the solution satisfies the initial condition.

Thus, we have shown that the solution to the differential equation is:

\[
i = I e^{-\frac{R}{L} t}
\]

Do you want more details or have any questions?

---

Here are some related questions to expand on this topic:

1. What is the physical interpretation of \(R\) and \(L\) in this equation?
2. How does the solution change if \(R = 0\)?
3. Can this equation be extended to circuits with capacitors (RL or RLC circuits)?
4. How do initial conditions affect the solution to a differential equation like this?
5. What is the time constant in this system, and how does it affect the decay of the current?

**Tip**: For differential equations involving circuits, initial conditions like \(i(0)\) are critical for determining the constants in the solution.

The current in an electric circuit is given by the equation Ri + L(di/dt) = 0, where L and R are constants. Show that i = Ie^(-Rt/L), given that i = I when t = 0.

This problem involves solving a first-order differential equation for the current in an RL circuit. Learn how to apply separation of variables and solve for the current using an exponential function. The problem is suitable for high school students studying Grades 11-12 physics or calculus.

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