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.