The given differential equation is:

\[
y' = 1 + x + y + xy
\]

This can be rewritten as:

\[
\frac{dy}{dx} = 1 + x + y + xy
\]

This equation is **not** separable or exact in its current form, but it can be simplified using a substitution technique. One common substitution is to use:

\[
v = y + 1
\]

Differentiating both sides with respect to \( x \), we get:

\[
\frac{dv}{dx} = \frac{dy}{dx}
\]

Now substitute \( y = v - 1 \) into the original equation:

\[
\frac{dv}{dx} = 1 + x + (v - 1) + x(v - 1)
\]

Simplifying:

\[
\frac{dv}{dx} = 1 + x + v - 1 + xv - x
\]

\[
\frac{dv}{dx} = v + xv
\]

Factor out \( v \):

\[
\frac{dv}{dx} = v(1 + x)
\]

This is a separable equation now. We can rewrite it as:

\[
\frac{dv}{v} = (1 + x) dx
\]

Integrating both sides:

\[
\ln|v| = \frac{x^2}{2} + x + C
\]

Exponentiating both sides:

\[
v = e^{\frac{x^2}{2} + x + C} = e^C e^{\frac{x^2}{2} + x}
\]

Let \( e^C = C_1 \), then:

\[
v = C_1 e^{\frac{x^2}{2} + x}
\]

Since \( v = y + 1 \), we have:

\[
y + 1 = C_1 e^{\frac{x^2}{2} + x}
\]

Finally, solving for \( y \):

\[
y = C_1 e^{\frac{x^2}{2} + x} - 1
\]

Thus, the general solution is:

\[
y = C_1 e^{\frac{x^2}{2} + x} - 1
\]

Would you like more details on any step or clarification?

### Related Questions:
1. How do you solve separable differential equations in general?
2. What are the conditions for a differential equation to be exact?
3. Can other substitution methods simplify this differential equation further?
4. How do boundary or initial conditions affect the constant \( C_1 \)?
5. What are common techniques to solve first-order differential equations?

### Tip:
Always check if a differential equation is separable, exact, or can be transformed via substitution for easier solving methods.

Learn how to solve the differential equation y' = 1 + x + y + xy using substitution methods to simplify the equation and find the general solution. This step-by-step guide explains the substitution technique and how to separate variables to integrate. Suitable for advanced calculus students.

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