is it relevent getting a solution of differential equation which is not a function by definition?

Question

simple example: $$2y \frac{dy}{dx}=1$$

solution is $$y^2=x+c$$ which is not a funtion in $x$ as it has two value of $y$ for one input $x$

Answer 1

To expand on Lorenzo Q's comment.

What you've found in $y^2 = x + c$ is not a "solution", but rather a condition on the function $y = f(x)$ which any solution of $$2y\frac{dy}{dx} = 1$$ must satisfy. In fact, with some basic qualifications (for example, continuity), it is a condition which is equivalent to the original differential equation.

However, as you've noted, it does not by itself define a function. So you have to go a step further to find functions that satisfy it. Fortunately, this is much simpler than for the differential equation. For any $x$, the corresponding $y$ must satisfy $$y = \pm\sqrt{x+c}$$ When solving $y^2 = x + c$, for each $x > -c$, we can choose whether $y$ should be $+\sqrt{x+c}$ or $-\sqrt{x+c}$ independently. So there are infinitely many (indeed, uncountably many) different functions that satisfy this condition.

But for the differential equation, we need $y$ to be differentiable, and therefore continuous. That only occurs if we pick $y = +\sqrt{x+c}$ for all $x$, or we pick $y = -\sqrt{x+c}$ for all $x$. So with that additional restriction, we have exactly two solutions.

Anytime you solve a differential equation (or more generally, any equation), you are actually converting your original problem into another problem whose answer is easier. When you go from $y'' + y = 0$ to $y = A\cos x + B\sin x$, you've simply transformed your original problem to one that is trivial. If you end up at some state where you don't have a definite solution, it just means you haven't finished solving yet.