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# Introduction

In this quiz, we’re going to take a close look at the solutions to one particular equation. We’re not concerned yet with the technique for finding the solutions (we’ll see how to solve it in Chapter 2). We’re more interested now in making qualitative statements about what the solutions to differential equations "look" like.

The equation is \[y^2 \; \cdot \; \frac{dy}{dx} = x^2.\] Which of the following is a solution to the equation?

I. \(y = \sqrt[3]{x^3 + 1}\)

II. \(y = \sqrt{x^3 + 1}\)

III. \(y = \sqrt[3]{x^3 - 10}\)

IV. \(y = x\)

*complete set* of solutions to \(y^2 \; \cdot \; \frac{dy}{dx} = x^2 ?\) \((\)The letter \(C\) denotes any real number.\()\)

The full solution to the differential equation \(y^2 \; \cdot \; \frac{dy}{dx} = x^2\) is \(y = \sqrt[3]{x^3 + C}.\) Observe that this is an infinite family of solutions, one for each choice of \(C.\) This is called a **one-parameter family** of solutions because we get all the solutions by adjusting the single parameter \(C.\) Note the similarities to finding indefinite integrals, where the answer is often in a form like \(y = \frac{1}{2}x^2 + C.\) But here the \(C\) can appear anywhere in the expression, not just added to the end.

It’s typical for a first-order differential equation, i.e. an equation that doesn’t involve higher-order derivatives, to have a one-parameter family of solutions. At the end of this quiz, we’ll see an example of a second-order equation whose solution is a two-parameter family.

There are two ideas we want to explore with this equation: integral curves and initial value conditions.

Suppose we graph some of the solutions to our equation. In other words, we choose some values of \(C,\) and for each choice, we graph \(y = \sqrt[3]{x^3 + C}\) on the same axes. We get a bunch of curves. What do these curves look like?

A.

B.

C.

D.

When a one-parameter family of solutions to a differential equation is graphed on the same axes, the curves are called **integral curves**. They resemble curves on a contour map, and we think of them as “filling out” the whole plane.

Each integral curve, of course, represents the solution corresponding to some specific value of \(C.\) You can often get global insights into the differential equation from looking at its integral curves.

Sometimes, as we will see, we won’t be able to solve the equation exactly, but we can still sketch its integral curves, and therefore get an understanding of how the solutions behave.

Now let's talk about initial value conditions. Consider these two statements about the solutions to \(y^2 \; \cdot \; \frac{dy}{dx} = x^2.\)

For every choice of \(a,\) there is a unique solution with the property that \(y(0) = a.\) (In other words, every point on the line \(x=0\) intersects exactly one integral curve.)

For every choice of \(a,\) there is a unique solution with the property that \(y(1) = a.\) (In other words, every point on the line \(x=1\) intersects exactly one integral curve.)

Which statements are true?

The last question graphically demonstrates an important point: typically with a one-parameter family of solutions, we can determine a particular solution by specifying its value at some point, like \(x = 0\) or \(x = 1.\) We use this information to determine the value of \(C\).

Since \(x\) often represents time, specifying a value at \(x = 0\) means specifying an *initial condition* of the equation.

What is the solution to \(y^2 \; \cdot \; \frac{dy}{dx} = x^2\) satisfying the initial condition \(y(0) = -2?\)

Let’s finish the quiz by looking at a more complicated example. Consider the equation
\[\frac{d^2y}{dt^2}= 1 - \frac{dy}{dt}.\]
This is an example of an equation we looked at in the last quiz: it describes how far an object falls through the air as a function of time \(t,\) taking air resistance into account. We will see in Chapter 3 how to solve this equation. Now we just state the general solution:
\[y(t) = t + C_1e^{-t} + C_2.\]
Observe that there are two constants: this is a *two-parameter* family of solutions. This corresponds to the fact that the original equation now involves the *second derivative*.

You might think that to determine a specific solution, you would need to specify two conditions, and you would be right! One way to do this is to specify the values of \(y\) and its derivative \(y'\) at 0. This corresponds here to specifying the initial position and initial velocity of the falling object.

What is the solution to this equation corresponding to \(y(0) = 0\) and \(y'(0) = 2?\)

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