Solve the equation: $$x^2 \cdot y' +2x\cdot y=1$$
Development:
This is a linear ODE, since dividing the equation by $$x$$ (note that $$x$$ cannot be a zero, since the equality would not be satisfied).
Therefore, the ODE can be re-written as: $$$y'=-\dfrac{2}{x}\cdot y+\dfrac{1}{x^2}=a(x)\cdot y+b(x)$$$ and clearly, that is a linear ODE.
Homogeneous ODE:
We look for a solution of the homogeneous ODE: $$y'=-\dfrac{2}{x}\cdot y$$ which is a separable ODE that we can solve: $$$\dfrac{dy}{dx}=-\dfrac{2}{x}\cdot y \Rightarrow \dfrac{dy}{y}=-\dfrac{2}{x}dx \Rightarrow \int \dfrac{dy}{y}=-\int \dfrac{2}{x}dx \Rightarrow $$$ $$$ \Rightarrow \ln|y|=\ln|x^{-2}|+C \Rightarrow y(x)=\dfrac{k}{x^2}, \ k\in\mathbb{R}$$$
Non homogeneous ODE:
We look for a particular solution of the type $$y_p(x)=u(x)\cdot y_1(x)$$ where $$y_1(x)=\dfrac{1}{x^2}$$. We know that the function $$u(x)$$ is a solution of $$$u'=\dfrac{b(x)}{y_1}=\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}}=1$$$ therefore $$u(x)=x$$.
And so, a particular solution is: $$$y_p(x)=u(x)\cdot y_1(x)=x\cdot\dfrac{1}{x^2}=\dfrac{1}{x}$$$ Finally, the solution will be the sum: $$$y(x)=y_h(x)+y_p(x)=\dfrac{k}{x^2}+\dfrac{1}{x}, \ k\in\mathbb{R}$$$
Solution:
$$y(x)=\dfrac{k}{x^2}+\dfrac{1}{x}, \ k\in\mathbb{R}$$