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The image contains three mathematical questions. Let's break them down one by one with detailed explanations:

### 1. **Differential Equation:**
   \[
   \text{Résoudre l'équation différentielle: } y' = 1 + y^2.
   \]

   This is a first-order non-linear differential equation of the form:
   \[
   \frac{dy}{dx} = 1 + y^2.
   \]
   - **Step 1: Separation of Variables**
     To solve this, we use the method of separation of variables. Rewrite the equation as:
     \[
     \frac{dy}{1 + y^2} = dx.
     \]
   - **Step 2: Integrate Both Sides**
     We integrate both sides:
     \[
     \int \frac{1}{1 + y^2} dy = \int 1 dx.
     \]
     The integral of \( \frac{1}{1 + y^2} \) is \( \arctan(y) \), and the integral of 1 is \( x + C \), where \( C \) is the constant of integration:
     \[
     \arctan(y) = x + C.
     \]
   - **Step 3: Solve for \( y \)**
     To find \( y \), apply the inverse of \( \arctan \):
     \[
     y = \tan(x + C).
     \]
     Thus, the general solution to the differential equation is:
     \[
     y = \tan(x + C).
     \]

### 2. **Solving a Second-Order Differential Equation:**
   \[
   \text{Déterminer la solution de l'équation } x^2y'' + 3xy' + y = 0.
   \]
   This is a Cauchy-Euler differential equation, which has the general form:
   \[
   x^2 y'' + axy' + by = 0.
   \]
   Here, \( a = 3 \) and \( b = 1 \).

   - **Step 1: Assume a Solution**
     We assume a solution of the form:
     \[
     y(x) = x^r.
     \]
     - First derivative: \( y'(x) = r x^{r-1} \)
     - Second derivative: \( y''(x) = r(r-1) x^{r-2} \)

   - **Step 2: Substitute into the Differential Equation**
     Substitute \( y(x) = x^r \) and its derivatives into the equation:
     \[
     x^2[r(r-1)x^{r-2}] + 3x[r x^{r-1}] + x^r = 0.
     \]
     Simplifying each term:
     \[
     r(r-1) x^r + 3r x^r + x^r = 0.
     \]
     Factor out \( x^r \):
     \[
     x^r[r(r-1) + 3r + 1] = 0.
     \]
     This gives the characteristic equation:
     \[
     r(r-1) + 3r + 1 = 0.
     \]
     Simplifying:
     \[
     r^2 + 2r + 1 = 0.
     \]
     This is a perfect square:
     \[
     (r + 1)^2 = 0.
     \]
     So, \( r = -1 \).

   - **Step 3: General Solution**
     Since \( r = -1 \) is a repeated root, the general solution for a repeated root is:
     \[
     y(x) = C_1 x^{-1} + C_2 x^{-1} \ln(x).
     \]

### 3. **Functional Equation:**
   \[
   \text{Résoudre dans } \mathbb{R} : f'(x) + \frac{1}{x^2 + 1} f(x) = \frac{x}{\sqrt{x^2 + 1}}.
   \]
   This is a first-order linear differential equation. The general form of a first-order linear equation is:
   \[
   f'(x) + P(x) f(x) = Q(x),
   \]
   where \( P(x) = \frac{1}{x^2 + 1} \) and \( Q(x) = \frac{x}{\sqrt{x^2 + 1}} \).

   - **Step 1: Find the Integrating Factor**
     The integrating factor is given by:
     \[
     \mu(x) = e^{\int P(x) dx} = e^{\int \frac{1}{x^2 + 1} dx}.
     \]
     The integral of \( \frac{1}{x^2 + 1} \) is \( \arctan(x) \), so the integrating factor is:
     \[
     \mu(x) = e^{\arctan(x)}.
     \]
   - **Step 2: Multiply the Equation by the Integrating Factor**
     Multiply the entire differential equation by \( \mu(x) = e^{\arctan(x)} \):
     \[
     e^{\arctan(x)} f'(x) + e^{\arctan(x)} \frac{1}{x^2 + 1} f(x) = e^{\arctan(x)} \frac{x}{\sqrt{x^2 + 1}}.
     \]
     The left-hand side becomes:
     \[
     \frac{d}{dx} \left( e^{\arctan(x)} f(x) \right) = e^{\arctan(x)} \frac{x}{\sqrt{x^2 + 1}}.
     \]

   - **Step 3: Integrate Both Sides**
     Integrate both sides with respect to \( x \):
     \[
     e^{\arctan(x)} f(x) = \int e^{\arctan(x)} \frac{x}{\sqrt{x^2 + 1}} dx.
     \]
     Solving this integral and isolating \( f(x) \) would give the final solution, but this is where the algebraic complexity lies. The general approach involves advanced integration techniques.

---

### Follow-Up Questions:
1. Can you provide the specific method to compute integrals involving functions like \( e^{\arctan(x)} \)?
2. How do we handle cases where the characteristic equation in second-order differential equations has complex roots?
3. What are some common applications of Cauchy-Euler equations in physics or engineering?
4. Could you explain the meaning of the integrating factor in a more intuitive way?
5. What are the conditions under which the solution to a differential equation is unique?

### Tip:
When solving linear differential equations, the integrating factor method is a powerful tool to reduce the equation to an easily solvable form. Always ensure that the integrating factor is correctly computed, as it simplifies the solution process dramatically.

This page provides a detailed step-by-step solution to three differential equations. Learn how to solve a first-order non-linear equation, a Cauchy-Euler equation, and a first-order linear equation using the integrating factor method. Suitable for university students studying advanced calculus.

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