difference between implicit and explicit solutions? What is difference between implicit and explicit solution of an initial value problem?
Please explain with example both solutions(implicit and explicit)of same initial value problem? Or without example but in some way that is understandable.

thanks
As requested:

Let's use the example initial-value problem

$$y^\prime y=-x,\qquad y(0)=r, \qquad r\text{ constant}$$

One can derive both an implicit and explicit solution for this DE. The implicit solution to this DE is

$$x^2+y(x)^2=r^2$$

This solution implicitly defines $y(x)$; all we have here is an equation involving $y(x)$. On the other hand, the explicit solution looks like

$$y(x)=\pm\sqrt{r^2-x^2}$$

and in this case, $y(x)$ is explicitly defined: $y(x)$ is expressed here as an explicit function with $x$ as the only independent variable.



We aren't always this lucky when we solve differential equations that show up in practice. It often happens that we can only be content with an implicit solution (or a parametric solution, which is a somewhat better state of affairs than having just an implicit solution). One famous example is the differential equation that pops up in the brachistochrone problem:

$$(1+(y^\prime)^2)y=r^2$$
Let us consider a differential equation
$$
x +yy' =0 \label{1}\tag{1}
$$
Now, the relation $x²+y² -25 =0$ is an implicit solution of the above equation for all $x\in(-5,5)$. This relation define two real functions $f_1(x) =\sqrt{25 -x^2}$ and $f_2(x) = -\sqrt{25 -x²}$ in the interval $x\in(-5,5)$: here $f_1,f_2$ are explicit solutions of the differential equation.

Now is it quite easy to understand that what are implicit and explicit solutions?


  Implicit solution means a solution in which dependent variable is not separated and explicit means dependent variable is separated.


Now consider the relation
$$
x² +y² +25 =0  
$$
Is it also an implicit solution of the differential equation \eqref{1}?
The answer is 'No': formally it is solution of the equation \eqref{1} but not implicitly because it won't be identically zero for any real values of $x,y$. Precisely If you write it in the form
$$
y  = -\sqrt{-25 -x^2}\:\text{}.
$$
you will get always an imaginary $y$ for all real $x$.