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Mathematics · Ordinary differential equations · 02

Solution Families and Initial Conditions

One rule of change can leave endless curves. How does a single initial condition pick out one of them?

Staying with y′ = y, find the general solution from the family of solutions, then put a chosen starting point into the formula, one step at a time.

Series · 02 / 03

Place one point inside a whole family of solutions.

The equation is still y′ = y. Change the starting point and see which curve passes through it.

Chosen exact solutionInitial pointOther solutions (faint dashes)
The solutions of y′=y are Ceˣ; y(x₀)=y₀ picks out y₀e^(x−x₀).
Tap the plot, use the sliders, or use the arrow keys to move the initial point. x₀ can go from −0.5 to 0.5 and y₀ from −2 to 2. A tap outside that range stops at the nearest edge.
Take this point to the slope field →
Initial point (x₀, y₀)(0, 1)
C = y₀e−x₀ (approx.)1
This solution at x = 1 (approx.)2.718

01 / Picking up from “Differential Equations”

One rule, many possible starting points.

For y′ = y, eˣ multiplied by any constant is still a solution. That whole group of curves is called a family of solutions.

In “Differential Equations” you checked eˣ, 3eˣ and 0, and all of them satisfy y′=y. Now write the different multiples with a single symbol, C.

y(x) = CeˣC is any real constant. Choose a C, and you have chosen one function.

With C fixed, differentiate with respect to x and you get y′=Ceˣ=y. So every C gives a solution, and together they form a family of solutions.

When C is positive, the curve lies above the x-axis. When C is negative, it lies below. When C is 0, it is the flat zero solution. The faint curves in the plot show only a few members of this infinite family; the green curve is the one you have chosen.

C is one fixed number for the whole curve. As you move along x, you can’t pick a new C at each point. If you replace it with C(x), the derivative gains an extra term C′(x)eˣ, and usually the function no longer satisfies the equation.

02 / More than a few examples

How do you know no solution is missing?

Saying “these are all solutions” is not enough. You also need to say “there are no others.” This section uses one small trick to force any solution into the same form.

Checking that every Ceˣ is a solution does not prove that every solution looks like that. For that, you go the other way: assume y is any differentiable solution, and see what form the equation forces on it.

  1. Start from the equation. On the connected interval you are considering, y′−y is 0 everywhere.

    y′ − y = 0

  2. Look at the product of y and e⁻ˣ. Differentiate it with the product rule and the chain rule.

    (e⁻ˣy)′ = −e⁻ˣy + e⁻ˣy′

  3. Factor out e⁻ˣ and use the equation.

    (e⁻ˣy)′ = e⁻ˣ(y′−y) = 0

  4. A derivative that is 0 everywhere means the function is constant on that interval. This is a calculus result that follows from the mean value theorem.

    e⁻ˣy = C

  5. Multiply by eˣ to solve for y. eˣ is never 0, so this step cannot lose the zero solution.

    y = Ceˣ

So Ceˣ covers every differentiable solution of this equation on a connected interval. It is called the general solution. In this example, the formula extends to the whole real line.

Refresher: why does the product rule give two terms?

If u and v both change with x, the derivative of their product is (uv)′=u′v+uv′. You can’t differentiate just one of the factors.

Here u=e⁻ˣ and v=y. The chain rule gives u′=−e⁻ˣ, so the two terms combine into exactly e⁻ˣ(y′−y). This idea of multiplying by e⁻ˣ will grow into the integrating factor later on.

03 / Naming a point the curve must pass

The equation gives the rule; the initial condition gives the place.

The equation says how the curve changes. The initial condition says where it starts. For y′ = y, the two together pick out exactly one curve.

y(0)=2 means that at x=0 the function’s value must be 2. On the graph, the curve must pass through the point (0,2).

Put the differential equation and this condition together, and you have an initial value problem:

y′ = y,　y(0) = 2

  1. Write the general solution.

    y(x) = Ceˣ

  2. Substitute x=0. y(0) is the function’s value at 0, not y times 0.

    y(0) = Ce⁰ = C

  3. Require that value to be 2.

    C = 2

  4. Put the constant back into the function.

    y(x) = 2eˣ

Once its constant is fixed, this solution can be called the particular solution of the initial value problem. It has to pass two checks: it follows the rule of change, and it passes through the given starting point.

Check the equation

y′=2eˣ=y
Holds for every x.

Check the initial condition

y(0)=2e⁰=2
Passes through the given point (0,2).

For example, 3eˣ also satisfies the equation, but it equals 3 at x=0, so it is not a solution of this initial value problem.

04 / The same calculation, in general

Leave the starting position as a symbol too.

The start doesn’t have to be at x = 0. Write it as the symbols (x₀, y₀), and one formula covers every starting point.

An initial condition can be given at any x. Let x₀ be the known position and y₀ the height there, and write y(x₀)=y₀. The subscript 0 marks initial data; it does not mean either value is zero.

  1. Substitute the starting position into the general solution.

    y₀ = Cex₀

  2. ex₀ is never zero, so you can divide by it.

    C = y₀e−x₀

  3. Put C back into y=Ceˣ and combine the exponents.

    y(x) = y₀ex−x₀

x−x₀ measures how far you have moved from the starting position. Set x back to x₀, the exponent becomes 0, and the function returns to y₀ on its own.

A full example: y(1)=−2

Here x₀=1 and y₀=−2, so C=−2/e, which gives:

y(x) = −2ex−1

When you differentiate, x−1 has derivative 1, so y′=−2ex−1=y. Substituting x=1 gives y(1)=−2e⁰=−2. Both checks pass.

This starting position lies outside the range of the sliders, yet the formula works just the same. The slider range is a display choice; it does not restrict the formula.

Back in the plot, first make y₀ negative and see where the curve sits. Then set it to 0. Finally, keep y₀ fixed and change only x₀, and watch how C adjusts so that the curve always passes through the starting point.

05 / How far the conclusion reaches

Unique here does not mean unique everywhere.

This problem has exactly one answer because we proved it, not because one constant and one condition automatically guarantee it.

For y′=y, you have proved that every solution is Ceˣ, and y(x₀)=y₀ fixes exactly one value, C=y₀e−x₀. So for any real starting point, this problem has exactly one solution on the whole real line.

That is a proof for this example. For differential equations in general, you have to check separate conditions for existence and uniqueness. Counting one constant and one condition guarantees nothing.

One initial condition, still two ways to go

Take y′=3y2/3, where y2/3=(∛y)², and require y(0)=0.

For y=0, both sides are 0. For y=x³, the derivative is 3x², and the right side is 3(∛x³)²=3x². Both functions pass through (0,0), yet they are different solutions.

So naming one starting point does not, by itself, guarantee uniqueness. Near zero, the right side of this equation does not meet the local Lipschitz condition in y that the usual uniqueness theorem needs. Later topics cover that condition and the theorem in detail.

“Particular solution” will turn up with a second meaning

Here it means the one solution in the general solution whose constant has been chosen to fit the initial condition. Later, for nonhomogeneous linear equations, “particular solution” means any one solution of the nonhomogeneous equation, found first, which need not meet a given initial value yet. The context will make clear which meaning is in play.

06 / Check the equation and the starting value separately

Work out the constant, then check the whole function.

First find the constant C. Then put the whole function back in and check it. Do both steps.

1. y′=y with y(0)=−1. What is the solution?

See the full reasoning

From y=Ceˣ, y(0)=C=−1, so y=−eˣ. Its derivative is −eˣ, which satisfies the equation, and its value at 0 is −1, which meets the initial condition.

2. y′=y with y(1)=2. Can you simply write y=2eˣ?

See the full reasoning

No. Substituting x=1 into the general solution gives Ce=2, so C=2/e and the solution is y=2ex−1. If you write 2eˣ, you get 2e at x=1, not the required 2.

3. If the starting height is 0, will the solution ever leave zero by itself?

See the full reasoning

For this page’s y′=y, no. Every x₀ gives C=0, and the only solution is y=0. That conclusion depends on this particular equation. The non-uniqueness example in the previous section is a reminder not to apply it to every equation.

You know the rule, and you have chosen a starting point. “Slope Fields and Euler’s Method” sets the full formula aside and looks only at the slope at each point: can local directions lead you to a whole curve?

Check and go further

Sources and scope

These three pages are about ordinary differential equations with one independent variable, and they use dimensionless math examples. The plot shows only a finite window. A curve leaving the window does not mean the solution ends there.

The pictures help you see; substitution is what checks that an example satisfies the equation. The general theory of existence and uniqueness, and more ways to solve equations exactly, come in later topics.