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.


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    $\begingroup$ Before anything else: from where did you encounter these terms? (I'm trying to gauge how to properly answer your question to suit you.) $\endgroup$ Oct 29, 2011 at 11:03
  • $\begingroup$ 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? $\endgroup$
    – Hafiz
    Oct 29, 2011 at 11:21
  • $\begingroup$ these are common terms in differential equations $\endgroup$
    – Hafiz
    Oct 29, 2011 at 11:22
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    $\begingroup$ 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 $y(x)-\varepsilon\sin(y(x))=x$... $\endgroup$ Oct 29, 2011 at 11:30
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    $\begingroup$ Sometimes you solve an differential equation, and the answer is something of the type : $y+x=\sin(xy)$. While you still don't know exactly (i.e. explicitly) what $y$ is, this relation usually yields enough information to answer to many questions about $y$. If you can find the solution as $y=f(x)$, that is always best, but if you cannot, an equation between $x,y$ is still much better than nothing. $\endgroup$
    – N. S.
    Oct 29, 2011 at 17:20

3 Answers 3


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


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


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:


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    $\begingroup$ I am wondering what the benefit of an implicit function is... If we know that the solution of an ODE has to satisfy some equation $f(x,y)=0$, what is then the advantage? $\endgroup$
    – asd
    Oct 28, 2015 at 10:23
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    $\begingroup$ @asd, because in some ways normal equations are easier to analyze than differential equations. $\endgroup$ Oct 22, 2017 at 13:12

Explicit solution is a solution where the dependent variable can be separated. For example, $x+2y=0$ is explicit because if y is dependent, I can rewrite it as $y=-\frac{x}{2}$ and my y has been separated.

Implicit is when the dependent variable cannot be separated like $\sin(x+e^y)=3y$.

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    $\begingroup$ Strictly speaking (even as mentioned in the accepted answer) $y+2y=0$ is an implicit solution, which can be converted into the explicit form $y=-\frac{x}{2}$ $\endgroup$
    – Anubis
    Oct 31, 2017 at 21:53
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    $\begingroup$ @Anubis You might want to edit $y + 2y$ to $x + 2y$ $\endgroup$
    – Willem
    Oct 19, 2019 at 14:15
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    $\begingroup$ @Willem Oh yes, thanks. But unfortunately I'm not allowed to edit it now. So, for anyone got confused, y+2y=0 should be x+2y=0 above. $\endgroup$
    – Anubis
    Oct 19, 2019 at 14:54

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 $x$.


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