https://file.aiquickdraw.com/mathbot/file/1724570475644ycek08gq.jpg

solve this above by non separable method and metion formula used in it and use more steps

Let's analyze and solve the differential equations in the image you provided. Since the request is to use a non-separable method, I'll employ techniques such as substitution or integrating factors, depending on the equation. I'll solve a couple of examples from the image to demonstrate.

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

This equation is non-separable and requires a substitution method.

Let \( u = x + y \). Therefore, \( \frac{du}{dx} = 1 + \frac{dy}{dx} \).

Rewriting the original equation:
\[
\frac{dy}{dx} = 1 + e^{u}
\]
Since \( \frac{dy}{dx} = \frac{du}{dx} - 1 \):
\[
\frac{du}{dx} - 1 = 1 + e^{u}
\]
\[
\frac{du}{dx} = 2 + e^{u}
\]

This is a separable equation in \( u \). Now, separate and integrate:
\[
\frac{du}{2 + e^{u}} = dx
\]

To integrate the left side, let's perform a substitution:
Let \( v = e^u \), so \( dv = e^u du \):
\[
\frac{dv}{v(2+v)} = dx
\]

Partial fraction decomposition:
\[
\frac{1}{v(2+v)} = \frac{A}{v} + \frac{B}{2+v}
\]
Solving for \( A \) and \( B \):
\[
1 = A(2+v) + Bv \implies 1 = 2A + Av + Bv
\]
\[
A + B = 0, \quad 2A = 1 \implies A = \frac{1}{2}, \quad B = -\frac{1}{2}
\]
So:
\[
\frac{1}{v(2+v)} = \frac{1/2}{v} - \frac{1/2}{2+v}
\]
Now, integrate:
\[
\frac{1}{2} \int \frac{dv}{v} - \frac{1}{2} \int \frac{dv}{2+v} = \int dx
\]
\[
\frac{1}{2} \ln |v| - \frac{1}{2} \ln |2+v| = x + C
\]
\[
\frac{1}{2} \ln \left|\frac{v}{2+v}\right| = x + C
\]
\[
\ln \left|\frac{v}{2+v}\right| = 2(x + C)
\]
\[
\frac{v}{2+v} = e^{2(x+C)}
\]
Back-substitute \( v = e^u = e^{x+y} \):
\[
\frac{e^{x+y}}{2 + e^{x+y}} = Ce^{2x}
\]
Simplifying, we get the final solution:
\[
e^{x+y} = \frac{2Ce^{2x}}{1 - Ce^{2x}}
\]
This can be rewritten in terms of \( y \) if needed, or left as is for the general solution.

### Problem 20
Given:
\[
\frac{dy}{dx} = \frac{1}{y-4x-1}^2
\]
**Solution:**

This equation can also be tackled using a substitution method. Let's set \( u = y - 4x - 1 \), so \( \frac{du}{dx} = \frac{dy}{dx} - 4 \).

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

This is a first-order differential equation. We can rewrite it as:
\[
\frac{du}{dx} = 4 + \frac{1}{u^2}
\]
Or equivalently:
\[
\frac{du}{dx} = \frac{4u^2 + 1}{u^2}
\]

Now let's integrate:
\[
\int u^2 \, du = \int (4u^2 + 1) \, dx
\]
Integrating both sides:
\[
\frac{u^3}{3} = 4x + C
\]
\[
u^3 = 12x + 3C
\]

Substituting back \( u = y - 4x - 1 \):
\[
(y - 4x - 1)^3 = 12x + 3C
\]

This is the general solution to the differential equation.

### Summary of the Methods:
1. **Substitution:** Identifying a substitution (like \( u = x + y \)) can simplify the equation into a more familiar form.
2. **Separation of Variables:** Once the equation is reduced, separating and integrating both sides.
3. **Partial Fraction Decomposition:** When faced with a complex fraction, breaking it down into simpler fractions that can be integrated individually.

Would you like a detailed explanation of any of these steps, or perhaps another example solved? Here are some related questions for further practice:

1. Solve \( \frac{dy}{dx} = y + e^{x+y} \) using substitution.
2. Find the general solution of \( \frac{dy}{dx} = x + y \).
3. Solve \( \frac{dy}{dx} = \frac{1}{y - x} \) using substitution.
4. What is the method of solving a Bernoulli differential equation?
5. Solve \( \frac{dy}{dx} = \frac{1}{(y-2x-3)^2} \) using an appropriate method.

**Tip:** When faced with a complex differential equation, check if a substitution might simplify the equation before applying more advanced techniques.

Explore how to solve non-separable differential equations using advanced mathematical techniques such as substitution, integrating factors, and partial fraction decomposition. Detailed examples and step-by-step solutions included.

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