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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

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Before anything else: from where did you encounter these terms? (I'm trying to gauge how to properly answer your question to suit you.) –  J. M. Oct 29 '11 at 11:03
    
during my assignment, I was making solution and I encounter these terms, I saw that overall these two are similar then where is difference? Is it just matter of writting the final solution in different way or actually procedure for finding implicit or explicit solution is different? –  Hafiz Oct 29 '11 at 11:21
    
these are common terms in differential equations –  Hafiz Oct 29 '11 at 11:22
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Okay. Take for instance the differential equation $y^\prime y=-x$ with initial condition $y(0)=r$. The implicit solution of this differential equation is $x^2+y(x)^2=r^2$; here $y(x)$ is implicitly defined. The explicit solutions look like $y(x)=\pm\sqrt{r^2-x^2}$; the solution is "explicit" in that the expression for $y(x)$ can entirely be expressed in terms of $x$. Here, we are lucky to get an explicit solution since we know how to solve quadratics; it often happens that we can only be content with an implicitly expressed $y(x)$, like in the case of $y(x)-\varepsilon\sin(y(x))=x$... –  J. M. Oct 29 '11 at 11:30
    
If you write this solution in answer than I will be able to accept answer that will give you points. :) And thanks for your kind effort –  Hafiz Oct 29 '11 at 14:00

1 Answer 1

up vote 6 down vote accepted

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:

$$(1+(y^\prime)^2)y=r^2$$

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