Consider the following IVP

$\bullet $ $\;\;y'=2\sqrt{y}\;\;\;\; , y(0)=0$

The solution of the above IVP is given by ; $$ y(x) = \cases{0 & for $x \le x_0$\cr (x-x_0)^2 & for $x > x_0$}$$

where $x_0 \ge 0$.

According to my understanding of IVP's, their solutions are curves that satisfy the ODE and also pass through the point specified in the initial conditions. Hence they must satisfy the initial conditions (say),$\;\;y(x_0)=y_0$.

Now the above IVP has 2 family of solutions. But what confuses me is that,

$y(x)= (x-x_0)^2$, for $x > x_0$ is a solution but,

$y(0)=(-x_0)^2$ which need not be $0$. (So it fails to satisfy the initial condition $y(0)=0$)

I feel I have misunderstood the concept.

I have the same doubt with solutions of the following IVP too,

$\bullet$ $\;\;y'=y^{1/3}\;\;\; , y(0)=0$

The solution of the above IVP is given by ;

$$y(x)=\cases{0 & for $x\le x_0$\cr \pm \left(\frac{2}{3}(x-x_0)\right)^{3/2} & for $x> x_0$}$$

where $x_0 \ge 0$.

Can anyone clear this misunderstanding of mine.

And also providing a geometrical interpretation of IVP would help a lot.


2 Answers 2


You are interpreting this formula wrong. It defines a piecewise defined function which is zero on $[0,x_0]$ and quadratic thereafter. A compact expression is $$y(x)=\max(0,x-x_0)^2.$$

Depending on convention, the solution that is zero on $[0,\infty)$ is missing in this construction or captured by $x_0=\infty$.

  • $\begingroup$ It was a really silly error that I made in interpreting the function. This answer clears my doubt. Thank you. $\endgroup$
    – Naive
    Commented May 30, 2017 at 7:56

The solution to a differential equation are curves that have different y-axis intercepts, I.e. family of curves that are shifted vertically.

You can think about the graph of any function $f$ and even if you shift it vertically, suppose by a constant $c$ the slope pattern won't change. That is, it's derivative will be same as derivative of original function. It can also be proven algebraically, as derivative of $f(x)+c$ is $f'(x)$ and derivative of $f(x)$ is also $f'(x)$.

Hence, after solving differential equation, I.e. tracing back the graph from its derivative function, we get a family curves parallel to each other (I hope you understand what parallel means here). Therefore we add an unknown constant $c$ so that we don't miss any curve.

Now, if we're given the value of function at any fixed value of $x$, then we'll get a uniques curve. Therefore we determine the constant using that value.

For your question, note that we have solved differential equation, I.e. the function is derivable I.e. it is continuous.Therefore $y(0)=0$ for both the pieces of this function. This implies $x_0=0$.


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