Show that all solutions of $y'= \frac{xy+1}{x^2+1}$ are of the form $y=x+C\sqrt {1+x^2}$ without solving the ODE I think I'm supposed to use the existence and uniqueness theorem, but I'm not sure how. (because we're not allowed to actually solve the ODE)
Also, is it ok if I just differentiate y and plug it in the ODE?
Thanks!
 A: You need to show that ALL the solutions of the given differential equation are of the given form, which means that plugging it in is not enough.
You're given a differential problem of the form : 
$$y'=f(x,y)$$
In such cases, a solution exists if $f(x,y)$ is either continuous everywhere, or around a given initial value.
In your case, the function at the RHS is :
$$f(x,y) = \frac{xy+1}{x^2+1}$$
It's obvious that $f$ is continuous $\forall \space (x,y) \in \mathbb R^2$, thus a solution to the differential equation exists.
To show that the solution is unique, you have to show that the function $f(x,y)$ is Lipschitz or in other words, it's derivative with respect to $y$ is bounded.
Truly then : 
$$|f(x,y_2)-f(x,y_1)|=\bigg|\frac{xy_2+1}{x^2+1} - \frac{xy_1+1}{x^2+1}\bigg|= \bigg|\frac{x(y_2-y_1)}{x^2+1} \bigg|=\bigg|\frac{x}{x^2+1} \bigg||y_2-y_1|$$
The function :
$$f(x)=\frac{x}{x^2+1}$$
is bounded and independent of $y_1,y_2$, thus there exists one unique solution to the given differential equation problem.
Now, after showing that is unique, if plugging it in works, then all the solutions of the given differential problem are of the given form.
A: A general idea that might work (but I did not try): if $y$ is a solution, define $v$ by setting $y(x)-x=v(x) \sqrt{1+x^2}$. Now try to prove that $v$ must be constant, by using the equation for $y$.
A: Apparently the answer is supposed to be as follows:
Let $y$ be a solution to the ODE. Define a new function $g(x)= \frac{y-x}{\sqrt{1+x^2}}$.
We'll prove that $g'(x)=0$ for all x, this would mean that $g(x)$ is constant.
From the previous equation we can see that $y= x+g(x)\sqrt{1+x^2}$, and so:
$y'=1+g'(x)\sqrt{1+x^2}+\frac{xg(x)}{\sqrt{1+x^2}}$
$g'(x)\sqrt{1+x^2}=y'-1-\frac{xg(x)}{\sqrt{1+x^2}}=y'-\frac{xy+1}{1+x^2}=0$
$g'(x)=0$
