Uniqueness of solution for a specific ode

Consider the following ode, $xy'+y=1$, Determine whether this ode has a unique solution for $y(0)=b$ where $b$ is arbitrary, if it does, find the values of $b$.

My attempt: Solving the above ode, I get $xy=x+c$, where $c$ is a constant. This expression holds for all x and all y. So if $y(0)=b$ then $c=0$ which would imply $xb-x=x(b-1)=0$. Hence, either $x=0$ or $b=1$. Since $x$ is allowed to vary, I deduce that $b=1$. Since the ODE is linear, the solution, since it exists must be unique. Is there an alternative way to show this? Is my method correct?

Solution 1:

If $y$ is a solution that is defined at $x=0$ then setting $x=0$ we must have $0\cdot y'(0) + y(0) = 1$, or $y(0) = 1$. Hence the only value of $b$ for which this system has a solution (defined at $x=0$) is $b=1$.

Note that $\hat{y}(x) = 1$ is a solution, so we can consider $y-\hat{y}$ and look for non zero solutions of $xn'+n=0$ passing through $z(0) = 0$ to see if it is unique (obviously $x \mapsto 0$ is a solution).

Suppose $z$ is a solution, and consider the solution on an interval $[x_0,\infty)$ with $x_0>0$. It is straightforward to show that the system is Lipschitz and hence has a unique solution which we can easily solve to get $z(x) = z(x_0) {x_0 \over x}$. Hence the solution on $x>0$ is $z(x) = z(x_0) {x_0 \over x}$ and the only way we can have $z(0) = 0$ is if $z(x_0) = 0$, or $z=0$.

Hence the only solution of the system is $\hat{y}(x) = 1$ for all $x$, in which case the only valid $b$ is $b=1$.