Solve the nonlinear differential equation yy'' + (y')² = ky'

The given differential equation is $yy'' + (y')^2 = ky'$.
Notice that the left side of the equation is the derivative of the product $yy'$ with respect to $x$:

$$\frac{d}{dx}(yy') = y''y + y'y' = yy'' + (y')^2$$

So, the differential equation can be rewritten as:

$$\frac{d}{dx}(yy') = ky'$$

Integrate both sides of the equation with respect to $x$:

$$\int \frac{d}{dx}(yy') dx = \int ky' dx$$ $$yy' = k \int y' dx$$ $$yy' = ky + C_1$$

where $C_1$ is an arbitrary constant of integration.

Case 1: $k \neq 0$.
The equation is $y \frac{dy}{dx} = ky + C_1$.
Separate the variables:

$$\frac{y}{ky + C_1} dy = dx$$

Integrate both sides:

$$\int \frac{y}{ky + C_1} dy = \int dx$$

To integrate the left side, we use algebraic manipulation:

$$\int \frac{1}{k} \left(1 - \frac{C_1}{ky + C_1}\right) dy = x + C_2$$ $$\frac{1}{k} \left(y - C_1 \int \frac{1}{ky + C_1} dy\right) = x + C_2$$ $$\frac{1}{k} \left(y - \frac{C_1}{k} \ln|ky + C_1|\right) = x + C_2$$ $$\frac{y}{k} - \frac{C_1}{k^2} \ln|ky + C_1| = x + C_2$$

This is the general implicit solution for $k \neq 0$.
Additionally, if $ky + C_1 = 0$, then $y = -\frac{C_1}{k}$, which is a constant solution. Let $C = -\frac{C_1}{k}$. Then $y=C$ is a solution. Substituting $y=C$ into the original differential equation gives $C(0) + (0)^2 = k(0)$, which is $0=0$. This constant solution $y=C$ is a singular solution not covered by the implicit form due to the logarithm.

Case 2: $k = 0$.
If $k=0$, the equation from Step 2 becomes $yy' = C_1$.

$$y \frac{dy}{dx} = C_1$$

Separate the variables:

$$y dy = C_1 dx$$

Integrate both sides:

$$\int y dy = \int C_1 dx$$ $$\frac{1}{2} y^2 = C_1 x + C_2$$ $$y^2 = 2C_1 x + 2C_2$$

Let $A = 2C_1$ and $B = 2C_2$.

$$y^2 = Ax + B$$

This is the general implicit solution for $k=0$. The constant solution $y=C$ is covered by this form when $A=0$, giving $y^2=B$, so $y=\pm\sqrt{B}$.

The solutions are:
For $k \neq 0$:

$$\boxed{\frac{y}{k} - \frac{C_1}{k^2} \ln|ky + C_1| = x + C_2}$$

And the singular solution:

$$\boxed{y = C}$$

For $k = 0$:

$$\boxed{y^2 = Ax + B}$$