The given differential equation is:

\[
y' + \frac{y}{x-4} = 5(x-4) \cdot \sqrt{y}
\]

### Step-by-step Solution:

This is a **first-order non-linear** differential equation due to the \(\sqrt{y}\) term on the right-hand side.

#### 1. **Substitute for simplification**:
Let \( x - 4 = z \), so \( \frac{d}{dx} = \frac{d}{dz} \), and \( y' = \frac{dy}{dx} \) becomes \( \frac{dy}{dz} \).

Now, rewrite the equation in terms of \( z \):

\[
\frac{dy}{dz} + \frac{y}{z} = 5z \cdot \sqrt{y}
\]

#### 2. **Substitution for \( u = \sqrt{y} \)**:
Let \( u = \sqrt{y} \), so that \( y = u^2 \) and \( \frac{dy}{dz} = 2u \frac{du}{dz} \).

Substitute this into the equation:

\[
2u \frac{du}{dz} + \frac{u^2}{z} = 5zu
\]

#### 3. **Divide through by \( u \)** (assuming \( u \neq 0 \)):

\[
2 \frac{du}{dz} + \frac{u}{z} = 5z
\]

#### 4. **Simplify the equation**:

\[
2 \frac{du}{dz} = 5z - \frac{u}{z}
\]

This is a first-order linear differential equation in \( u \). We can solve this using an integrating factor.

The equation can be written as:

\[
\frac{du}{dz} + \frac{u}{2z} = \frac{5z}{2}
\]

#### 5. **Find the integrating factor**:
The integrating factor is:

\[
\mu(z) = e^{\int \frac{1}{2z} dz} = e^{\frac{1}{2} \ln z} = \sqrt{z}
\]

#### 6. **Multiply the equation by the integrating factor**:

\[
\sqrt{z} \frac{du}{dz} + \frac{u}{2\sqrt{z}} = \frac{5z^{3/2}}{2}
\]

This simplifies to:

\[
\frac{d}{dz}(u\sqrt{z}) = \frac{5z^{3/2}}{2}
\]

#### 7. **Integrate both sides**:

\[
u\sqrt{z} = \int \frac{5z^{3/2}}{2} dz
\]

The integral of \( z^{3/2} \) is:

\[
\int z^{3/2} dz = \frac{2}{5} z^{5/2}
\]

So:

\[
u\sqrt{z} = \frac{5}{2} \cdot \frac{2}{5} z^{5/2} + C = z^{5/2} + C
\]

#### 8. **Solve for \( u \)**:

\[
u = z^2 + \frac{C}{\sqrt{z}}
\]

#### 9. **Substitute back \( u = \sqrt{y} \) and \( z = x - 4 \)**:

\[
\sqrt{y} = (x - 4)^2 + \frac{C}{\sqrt{x - 4}}
\]

Squaring both sides:

\[
y = \left((x - 4)^2 + \frac{C}{\sqrt{x - 4}}\right)^2
\]

This is the general solution to the differential equation.

---

Would you like a detailed explanation on any of these steps?

### 5 Follow-Up Questions:
1. How do you find the integrating factor for a linear differential equation?
2. Why is the substitution \( u = \sqrt{y} \) useful in this problem?
3. What are the types of differential equations and how are they classified?
4. Can this equation be solved using a different method, such as separation of variables?
5. What is the significance of initial conditions in solving a differential equation?

### Tip:
When solving non-linear differential equations, try transformations or substitutions to simplify the equation into a form that's easier to solve, such as a linear equation.

Explore how to solve the first-order nonlinear differential equation y' + y/(x-4) = 5(x-4)√y through substitution methods and integrating factors. This solution is suited for students studying Calculus II or Differential Equations at the undergraduate level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation