Find all solutions of differential equation $xy'+(1-x)y=0$ and $xy'+(1-x)y=1$. Also does there exist a solution such that $y(0)=0$ and $y(0)=1$ for both these D.Es ?
My attempt: $xy'+ (1-x)y=0 \implies$ $\int \frac{dy}{y} = \int \frac{x-1}{x} dx$ for $x\ne 0$ and $y(x)\ne 0$ for any $x \in \mathbb{R}$. Hence $y=\frac{k}{x}e^x$ for constant $k\in \mathbb{R}-\{0\}$. But $y(x)=0 \,\forall\, x$ is also a valid solution of this D.E. Hence there exist a solution such that $y(0)=0$. All possible solutions of this D.E are:
$$y=\left \{ \begin{array}{ll} \frac{k}{x}e^x & \mbox{if } x \ne 0 \\ 0 & \mbox{if } x =0 \end{array} \right. $$ where $k\in \mathbb{R}$. But $y(0)=1$ is not possible since $0\cdot \frac{dy}{dx}\mid_{x=0}+(1-0)y(0)=0$ yields $y(0)=0$. I am not sure wether this proves the first part of the question.
For the second part, we have the D.E $xy'+(1-x)y=1 \implies$ $\int d(xe^{-x}y) = \int e^{-x}dx$. Hence $xy=-1+ce^{x}$ where $c \in \mathbb{R}$. In this case, all possible solutions are:
$$y=\left \{ \begin{array}{ll} \frac{-1}{x}+\frac{c}{x}e^x & \mbox{if } x \ne 0 \\ 1 & \mbox{if } x =0 \end{array} \right. $$ $y(0)=0$ is not possible since $0\cdot \frac{dy}{dx}\mid_{x=0}+(1-0)y(0)=1$ yields $y(0)=1$. From the definition of $y$ above, is that a valid solution such that $y(0)=1$ ? Is this proof correct ?
Here is the correct solution for the first DE : as you have shown any solution of the DE must satisfy $y=\frac {ke^{x}} x$ for $x>0$ where $k$ is a constant. This function does not have a limit as $x \to 0$ unless $k=0$. Hence the condition $y \equiv 0$. There is no solution with $y(0)=1$.
Similarly, the second equation has a solution with $y(0)=1$ (corresponding to $c=1$) and it has no solution with $y(0)=0$.