Find a general solution y(x) to the differential equation y′=yx+2x2

${\displaystyle \frac{dy}{dx}=\frac{y}{x}+2x^2}$
${\displaystyle \frac{dy}{dx}-\frac{y}{x}=2x^2}$
(check linear form of differential equation)
${\displaystyle P=-\frac{1}{x},Q=2x^2}$
${\displaystyle I\cdot F=e^{-\int pdx}=e^{-\int -\frac{1}{x}dx}}$
${\displaystyle =e^{\log x}=x}$
${\displaystyle \Rightarrow y\left(IF\right)=\int Q\left(IF\right)+C}$
${\displaystyle \Rightarrow yx=\int 2x^{3}dx+C}$
${\displaystyle \Rightarrow xy=\frac{x^4}{2}+C}$
${\displaystyle \Rightarrow y=\frac{x^3}{2}+\frac{c}{x}}$ - is the solution

Another solution?

Yes, here:
Solve the linear equation $(dy(x))/(dx)=2x^2+y(x)/x:$
Subtract $y(x)/x$ from both sides:
$(dy(x))/(dx)-y(x)/x=2x^2$
Let $\mu (x)=e^{(}int-1/xdx)=1/x$
Multiply both sides by $\mu (x):$
$((dy(x))/(dx))/x-y(x)/x^2=2x$
Substitute $-1/x^2=d/(dx)(1/x):$
$((dy(x))/(dx))/x+d/(dx)(1/x)y(x)=2x$
Apply the reverse product rule $f(dg)/(dx)+g(df)/(dx)=d/(dx)(fg)$ to the left-hand side:
$d/(dx)(y(x)/x)=2x$
Integrate both sides with respect to x:
$\int d/(dx)(y(x)/x)dx=\int 2xdx$
Evaluate the integrals:
$y(x)/x=x^2+c_1$, where $c_1$ is an arbitrary constant.
Divide both sides by $\mu (x)=1/x$
Answer:
$y(x)=x(x^2+c_1)$