Implicit Solutions and Solution Intervals

Explicit and implicit: two ways to write a solution

An answer can’t always be written as y = ⋯. Sometimes you can only give a relation between x and y first, and then pick out a piece of the curve that works as a function.

An explicit solution puts y alone on one side, such as y=√(4−x²). An implicit solution gives a relation between x and y, such as x²+y²=4. The relation describes a curve; you still have to pick out a differentiable function y(x) from it before you have the kind of solution we use for an ODE.

So “some equation holds” is not where checking a solution ends. The upper and lower halves of the circle both satisfy the same relation, yet the whole circle is not a single-valued function y(x): one x usually corresponds to two heights.

The variables on this page are dimensionless. The interval attached to an initial value problem is a connected open interval containing the starting position, on which the function is differentiable and the equation is defined and holds. “Largest” means the same solution cannot be extended any further. It has nothing to do with how big the plot window is.

Worked implicit example: a half circle from an equation

One relation can hold several pieces of solution. The initial condition decides which piece you are on, and the endpoints have to be stated too.

1. Keep the original restriction, then multiply by y

   y y′ = −x, y≠0

   After multiplying, remember that the original equation is undefined at y=0. Multiplying does not turn it into a different problem.

2. Recognize the chain rule, then integrate

   d(y²/2)/dx=−x ⇒ y²/2=−x²/2+C

   The right side integrates to −x²/2; y is a function that changes with x.

3. Use the starting value to fix the constant

   x²+y²=C₂; 0²+2²=4 ⇒ C₂=4

   This gives the implicit relation x²+y²=4.

4. Let the starting value choose the branch

   y=+√(4−x²), −2<x<2

   y(0)=2 chooses the plus sign; the minus sign gives −2, which does not meet the starting value. On the chosen interval y>0, so it cannot suddenly change sign.

5. Check with implicit differentiation

   2x+2y y′=0 ⇒ y′=−x/y

   This step still uses y≠0. You can also differentiate the positive square root directly to get −x/√(4−x²).

6. Check both endpoints

   x=±2 ⇒ y=0

   There the denominator of the equation is zero, and the explicit curve no longer has a finite slope. Neither endpoint can be included, so the largest interval is (−2,2).

If the starting value is y(0)=−2 instead, the constant is still 4, but the negative square root is chosen. The two branches cannot loop around at the endpoints to form one solution y(x): besides not being single-valued, the original equation is undefined where they would meet.

Keep the absolute value in the logarithm

Integrating 1/y gives ln|y|, and the absolute value matters: it handles the positive branch and the negative branch separately.

For y≠0, an antiderivative is ∫dy/y=ln|y|+C. Check it on each branch:

ln y allows only y>0 and would miss the negative solutions; ln|y| works on each side but still excludes y=0. Taking exponentials gives |y|=e^Ce^F(x), where the right side is always positive: e^C can never equal zero. For nonzero solutions the sign can be absorbed into A=±e^C, but the zero solution has to be put back from the original equation. You can’t claim it comes from some finite C.

On the previous page, A∈ℝ for y′=2xy combines three parts: positive solutions, negative solutions, and the zero solution checked separately. That every A gives a solution on the whole real line is special to that example, not true of every separable equation.

Worked interval example: y′=y², y(0)=1

The plot window does not decide where a solution holds. This solution really does end at x = 1; the plot is not just running out of room.

1. Keep the equilibrium solution first

   y=0

   It satisfies the equation but not the starting value 1. Near this starting value we can separate where y≠0.

2. Separate and integrate

   y′/y²=1 ⇒ −1/y=x+C

   Differentiating −1/y with respect to y gives 1/y², so no ln|y| appears here.

3. Put in the starting value

   −1/1=0+C ⇒ C=−1

   So −1/y=x−1, which gives y=1/(1−x).

4. Check the equation and the starting value

   y′=1/(1−x)²=y²; y(0)=1

   Valid for x≠1; the starting point lies to the left of the singularity.

5. Choose the connected interval containing 0

   I=(−∞,1)

   You can’t stop at “all x≠1,” because that set is two pieces that don’t connect.

6. Show it cannot be extended to the right

   x→1⁻ ⇒ y→+∞

   Any differentiable extension would have to be continuous, and so have a finite value y(1); blowing up rules that out.

On (1,∞) the same formula also satisfies the equation, but it is a separate, negative solution, apart from the one with y(0)=1. Calling the two pieces together the solution of this initial value problem would hide the fact that it cannot cross x=1.

With y(0)=−1 instead, the calculation gives y=−1/(1+x), whose largest interval is (−1,∞). Where the singularity lies is not fixed by the equation alone; it also depends on the starting value. y(0)=0 gives the zero solution, valid on the whole real line.

Check the equation, not just the last line

To check an answer, go back to the original equation. It is not enough that the last line is a true equation.

From intervals to overall behavior

A complete answer has four parts: the equation, the starting value, the branch chosen, and the interval where it holds.

From now on, every answer should come with the equation, the starting value, the branch and the interval of validity. On the next page, when the slope depends only on y, we will first use a phase line to see where solutions go, and then check with a formula.

Sources and scope

Definitions and the conditions of theorems were checked against the authors and university materials below. The explanations, worked examples and interactive plots here are written independently. Decimals on the plots are rounded displays; the mathematical conclusions rest on derivation and substitution. This page does not prove the general existence and uniqueness theorems it cites.

• MIT 18.022 · Implicit function theorem: a local function described by a continuously differentiable relation with a nonzero partial derivative.
• Jiří Lebl · Separable equations · Implicit solutions: implicit solutions, solutions lost in separation, and choosing a branch.
• Jiří Lebl · Existence and uniqueness: local solutions and where the original equation holds.