To solve 𝑦2 + 𝑦 − 𝑥𝑦′ = 0 

The given differential equation is:

\[
y^2 + y - xy' = 0
\]

where \(y' = \frac{dy}{dx}\). This is a first-order differential equation, and we will aim to solve it step by step.

### Step 1: Rearrange the equation
We can isolate \(y'\) by rearranging the equation as follows:

\[
xy' = y^2 + y
\]

\[
y' = \frac{y^2 + y}{x}
\]

This is a separable differential equation, meaning we can separate the variables \(y\) and \(x\) on opposite sides of the equation.

### Step 2: Separate the variables
We rewrite the equation as:

\[
\frac{dy}{y^2 + y} = \frac{dx}{x}
\]

### Step 3: Simplify the left-hand side
We can simplify the left-hand side using partial fraction decomposition. Start by factoring \(y^2 + y\):

\[
y^2 + y = y(y + 1)
\]

So, we want to decompose the following fraction:

\[
\frac{1}{y(y + 1)} = \frac{A}{y} + \frac{B}{y + 1}
\]

Multiplying both sides by \(y(y + 1)\) gives:

\[
1 = A(y + 1) + By
\]

Expanding the right-hand side:

\[
1 = A y + A + By
\]

Now, collect like terms:

\[
1 = (A + B)y + A
\]

This gives the system of equations:

\[
A + B = 0
\]
\[
A = 1
\]

From \(A = 1\), we find \(B = -1\). So, we can rewrite the left-hand side as:

\[
\frac{dy}{y(y + 1)} = \frac{1}{y} - \frac{1}{y + 1} \, dy
\]

### Step 4: Integrate both sides
Now, we integrate both sides:

\[
\int \left( \frac{1}{y} - \frac{1}{y+1} \right) dy = \int \frac{dx}{x}
\]

The integrals are straightforward:

\[
\ln|y| - \ln|y+1| = \ln|x| + C
\]

### Step 5: Simplify the result
Using logarithm properties, we combine the terms on the left-hand side:

\[
\ln \left( \frac{|y|}{|y+1|} \right) = \ln|x| + C
\]

Exponentiating both sides to eliminate the logarithms:

\[
\frac{|y|}{|y+1|} = Cx
\]

where \(C = e^C\) is a constant. Finally, we obtain the implicit solution:

\[
\frac{y}{y+1} = Cx
\]

### Final Answer:
The solution to the differential equation is:

\[
\frac{y}{y+1} = Cx
\]

where \(C\) is a constant of integration.

---

Do you want further details or have any questions? Here are five related questions to expand your understanding:

1. How would you solve this differential equation if it were non-separable?
2. Can this solution be expressed in terms of \(y\) alone?
3. How do we handle boundary conditions with this type of solution?
4. What if \(y\) had initial conditions, how would we determine \(C\)?
5. How do logarithmic properties help in simplifying solutions in differential equations?

**Tip:** For separable differential equations, always aim to get all terms involving \(y\) on one side and terms involving \(x\) on the other side.

Explore the solution to the first-order differential equation y^2 + y - xy' = 0 using separation of variables and integration techniques. Learn how to apply partial fractions and logarithmic simplifications for a step-by-step solution. This problem is ideal for university-level calculus students.

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