$$ y ^ { \prime } ( t ) + r y ( t ) = r - r \int y ( z ) d z , \quad y ( \cdot ) = 1 $$

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To solve the integro-differential equation:

$$
y'(t) + r y(t) = r - r \int y(z) \, dz,
$$

with the initial condition $$ y(\cdot) = 1 $$ (assuming this means $$ y(t_0) = 1 $$ for some initial time $$ t_0 $$), we can proceed as follows:

### Step 1: Transform the Integro-Differential Equation into a Differential Equation

First, let's differentiate both sides of the given equation with respect to $$ t $$:

$$
\frac{d}{dt}\left( y'(t) + r y(t) \right) = \frac{d}{dt}\left( r - r \int y(z) \, dz \right).
$$

This yields:

$$
y''(t) + r y'(t) = -r y(t).
$$

Rearranging terms, we obtain a **second-order linear homogeneous differential equation**:

$$
y''(t) + r y'(t) + r y(t) = 0.
$$

### Step 2: Solve the Characteristic Equation

The characteristic equation corresponding to the differential equation is:

$$
\lambda^2 + r \lambda + r = 0.
$$

Solving for $$ \lambda $$:

$$
\lambda = \frac{-r \pm \sqrt{r^2 - 4r}}{2}.
$$

The nature of the roots depends on the discriminant $$ \Delta = r^2 - 4r $$:

1. **Case 1: $$ r < 4 $$ (Complex Roots)**
   
   The roots are complex:
   
   $$
   \lambda = \frac{-r}{2} \pm i \frac{\sqrt{4r - r^2}}{2}.
   $$
   
   Thus, the general solution is:
   
   $$
   y(t) = e^{-\frac{r}{2} t} \left[ C_1 \cos\left( \frac{\sqrt{4r - r^2}}{2} t \right) + C_2 \sin\left( \frac{\sqrt{4r - r^2}}{2} t \right) \right],
   $$
   
   where $$ C_1 $$ and $$ C_2 $$ are constants determined by initial conditions.

2. **Case 2: $$ r = 4 $$ (Repeated Real Roots)**
   
   The roots are real and equal:
   
   $$
   \lambda = -2.
   $$
   
   Thus, the general solution is:
   
   $$
   y(t) = e^{-2t} (C_1 + C_2 t).
   $$

3. **Case 3: $$ r > 4 $$ (Distinct Real Roots)**
   
   The roots are real and distinct:
   
   $$
   \lambda = \frac{-r \pm \sqrt{r^2 - 4r}}{2}.
   $$
   
   Thus, the general solution is:
   
   $$
   y(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t},
   $$
   
   where $$ \lambda_1 $$ and $$ \lambda_2 $$ are the two distinct roots.

### Step 3: Apply Initial Conditions

Assuming the initial condition is $$ y(t_0) = 1 $$, you would substitute $$ t = t_0 $$ into the general solution and solve for the constants $$ C_1 $$ and $$ C_2 $$ based on any additional initial conditions provided (such as $$ y'(t_0) $$).

### Summary

The general solution to the given integro-differential equation is the solution to the second-order linear homogeneous differential equation derived above. Depending on the value of $$ r $$, the solution will involve exponentials, sines, and cosines.

**Final General Solution:**

After differentiating, the general solution satisfies y″ + r y′ + r y = 0. Thus,

y(t) = e^–rt⁄2 [ C₁ cos(½√(4r – r²) t) + C₂ sin(½√(4r – r²) t) ]

where C₁ and C₂ are constants determined by initial conditions.
The general solution to the equation $$ y'(t) + r y(t) = r - r \int y(z) \, dz $$ with the initial condition $$ y(t_0) = 1 $$ is: $$ y(t) = e^{-\frac{r}{2} t} \left[ C_1 \cos\left( \frac{\sqrt{4r - r^2}}{2} t \right) + C_2 \sin\left( \frac{\sqrt{4r - r^2}}{2} t \right) \right] $$ where $$ C_1 $$ and $$ C_2 $$ are constants determined by the initial conditions.

\( y ^ { \prime } ( t ) + r y ( t ) = r - r \int y ( z ) d z , \quad y ( \cdot ) = 1 \)

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