Recipe

Solve an initial-value problem and find its interval of validity

Use when: you're given a differential equation together with a starting value $y(x_0)=y_0$ and asked for the particular solution and how far it can be trusted

Goal: the particular solution passing through $(x_0,y_0)$ , plus the largest interval around $x_0$ over which it's actually valid

Taught in Initial-value problems

Quick recipe

Find the general solution
Apply $y(x_0)=y_0$ to solve for the constant
Find every value of $x$ where the particular solution breaks down
Keep only the piece of the domain containing $x_0$

1
Solve the differential equation by whatever method applies, keeping the arbitrary constant.

Example

For $\dfrac{dy}{dx}=-\dfrac{x}{y}$ : separating gives $y\,dy=-x\,dx$ , and integrating gives $\dfrac{y^2}{2}=-\dfrac{x^2}{2}+C$ , i.e. $x^2+y^2=2C$ .

Why: an IVP's particular solution is just one member of this whole family, picked out in Step 2.
2
Substitute the given point into the general solution and solve for the constant.

Example

With $y(2)=1$ : $2^2+1^2=2C$ , so $C=\dfrac52$ , giving $x^2+y^2=5$ . Since $y(2)=1>0$ , it's the positive branch $y=\sqrt{5-x^2}$ that actually passes through $(2,1)$ , not the negative one.

Why: this is what turns the whole family of solutions into the one specific curve through $(x_0,y_0)$ .
3
Look for a division by zero, a square root or logarithm of a negative or zero quantity, or any other place the formula stops being defined or differentiable.

Example

$y=\sqrt{5-x^2}$ requires $5-x^2\ge0$ , i.e. $-\sqrt5\le x\le\sqrt5$ . At the endpoints $y=0$ , and the original equation $\dfrac{dy}{dx}=-x/y$ divides by $y$ , so the solution isn't differentiable there either.
- a break point exists on one or both sides of $x_0$ — continue to Step 4 to select the correct interval.
- no break point exists anywhere — the solution is valid on the entire real line; Step 4 isn't needed.
Why: an interval of validity has to be a place the solution is BOTH defined and differentiable, not just where the formula avoids an outright error.
4
Among the break points found in Step 3, keep the largest open interval around $x_0$ containing none of them — that's the interval of validity, even if the full domain has other pieces too.

Example

$x_0=2$ lies strictly between $-\sqrt5\approx-2.236$ and $\sqrt5\approx2.236$ , with no break point in between, so $\boxed{y=\sqrt{5-x^2},\quad \text{valid on } (-\sqrt5,\ \sqrt5)}$ an interval bounded on BOTH sides of $x_0$ , unlike a break point that only cuts off one side.

Why: a break point on the far side of a different break point from $x_0$ says nothing about the solution near $x_0$ — the interval is built around $x_0$ , not around the equation as a whole.

Common mistakes

Reporting the domain of the general solution's formula instead of the particular solution actually passing through $(x_0,y_0)$ .
Stopping at "the formula is defined here" without also checking differentiability — a solution can be defined at a point where it still isn't differentiable.
Including a piece of the domain that sits on the far side of a break point from $x_0$ , when only the piece actually containing $x_0$ counts as the interval of validity.
Assuming a well-behaved-looking differential equation guarantees a solution valid for all $x$ , instead of checking the particular solution's own formula.