Let's tackle the differential equations you provided using non-separable methods.

### Problem 18
Given:
\[
\frac{dy}{dx} = 1 + e^{x+y+5}
\]

### Solution:
This differential equation is not separable in its current form. We'll use a substitution method to simplify it.

1. **Substitution:**

   Let \( u = x + y \). Then, differentiate both sides with respect to \( x \):
   \[
   \frac{du}{dx} = 1 + \frac{dy}{dx}
   \]

   Substitute this into the original equation:
   \[
   \frac{du}{dx} = 1 + 1 + e^{u+5}
   \]
   \[
   \frac{du}{dx} = 2 + e^{u+5}
   \]
   
2. **Simplification:**

   This is a first-order linear differential equation in \( u \). To solve it, rewrite the equation:
   \[
   \frac{du}{dx} - 2 = e^{u+5}
   \]

   To simplify further, let \( v = u + 5 \) so that \( \frac{dv}{dx} = \frac{du}{dx} \) and the equation becomes:
   \[
   \frac{dv}{dx} = 2 + e^{v}
   \]
   
3. **Solving the Equation:**

   Now, rearrange and integrate:
   \[
   \frac{1}{2 + e^{v}} dv = dx
   \]

   This integral may not have a simple analytical solution, but it can be approached by integrating numerically or using series expansions in certain cases. However, we can write the implicit solution as:
   \[
   \int \frac{dv}{2 + e^v} = x + C
   \]

   This expression represents the general solution, though it's not in a closed form.

### Problem 20
Given:
\[
\frac{dy}{dx} = \left(\frac{1}{y - 4x - 1}\right)^2
\]

### Solution:
Again, this is not separable, but we can solve it using substitution.

1. **Substitution:**

   Let \( u = y - 4x - 1 \). Then:
   \[
   \frac{du}{dx} = \frac{dy}{dx} - 4
   \]

   Substitute into the original equation:
   \[
   \frac{du}{dx} = \frac{1}{u^2} + 4
   \]

2. **Simplification and Solution:**

   The equation now becomes:
   \[
   \frac{du}{dx} = 4 + \frac{1}{u^2}
   \]

   Multiply through by \( u^2 \) to obtain:
   \[
   u^2 \frac{du}{dx} = 4u^2 + 1
   \]
   
   This can be written as:
   \[
   \frac{du}{4u^2 + 1} = dx
   \]

   Integrating both sides:
   \[
   \int \frac{du}{4u^2 + 1} = x + C
   \]

   The integral on the left side is standard and can be solved using the arctangent formula:
   \[
   \frac{1}{2} \arctan\left(\frac{2u}{1}\right) = x + C
   \]

   Solving for \( u \):
   \[
   u = \frac{1}{2} \tan(2x + 2C)
   \]

3. **Back Substitution:**

   Substituting back \( u = y - 4x - 1 \):
   \[
   y - 4x - 1 = \frac{1}{2} \tan(2x + 2C)
   \]

   Finally, the general solution is:
   \[
   y = 4x + 1 + \frac{1}{2} \tan(2x + 2C)
   \]

This represents the solution to the given differential equation.

### Summary:
- **Substitution** is a powerful technique for solving non-separable differential equations, especially when a direct separation of variables is impossible.
- These equations often lead to integrals that can be solved using standard methods like arctangent formulas.

Would you like further details on any of the steps, or perhaps another example solved? Here are some related questions for further exploration:

1. Solve \( \frac{dy}{dx} = y^2 + x^2 \) using a suitable method.
2. How do you solve a Riccati differential equation?
3. Apply the method of undetermined coefficients to solve a non-homogeneous linear differential equation.
4. What are the steps to solve an exact differential equation?
5. Solve \( \frac{dy}{dx} = e^{x+y} \) using substitution.

**Tip:** When encountering complex differential equations, look for substitutions that simplify the problem into a form that you can handle, such as a linear equation or an equation with standard integrals.

Explore how to solve differential equations using non-separable methods like substitution and integration techniques. This detailed guide covers specific examples and step-by-step solutions, including integration formulas such as integration by parts and the arctangent formula. Suitable for advanced mathematics students and enthusiasts.

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