Solutions, Initial Conditions, and Boundary Conditions, differential equations lesson

Which function has derivative $2x$ ? Every function of the form $x^2 + C$ does, one for each value of the constant, so the equation $y' = 2x$ with $y' = dy/dx$ describes a whole family of curves rather than a single one. Picking one member takes extra data, given either at one point or at the two ends of an interval, and the two kinds of data behave very differently. This lesson sorts out the kinds of solutions, shows how to verify a solution completely, and explains why every solution comes with an interval.

Key terms

Explicit and implicit solutions

Definition Let $y' = dy/dx$ . An explicit solution of an equation of order $n$ on an open interval $I$ is a function $y = \varphi(x)$ that is $n$ times differentiable on $I$ and satisfies the equation at every $x$ in $I$ . A relation $G(x, y) = 0$ is an implicit solution on $I$ if it defines at least one explicit solution on $I$ , that is, a function $\varphi$ with $G(x, \varphi(x)) = 0$ for every $x$ in $I$ that solves the equation there. A relation is checked by implicit differentiation, treating $y$ as a function of $x$ , so the chain rule brings in a factor $y'$ .

In plain words An explicit solution hands you $y$ as a formula in $x$ . An implicit one hands you a curve, and the solutions are the pieces of that curve that are graphs of differentiable functions.

Example The circle $x^2 + y^2 = 25$ is an implicit solution of $y' = -x/y$ on $(-5, 5)$ . Differentiating the relation gives $2x + 2yy' = 0$ , so $y' = -x/y$ wherever $y \ne 0$ . The full circle is not the graph of a function, because each $x$ in $(-5, 5)$ is the first coordinate of two points of the circle. It splits into two explicit solutions on $(-5, 5)$ , the upper and the lower half,

\begin{gathered} y = \sqrt{25 - x^2}, \\[0.4em] y = -\sqrt{25 - x^2}. \end{gathered}

For the upper half the derivative is $-x/\sqrt{25 - x^2}$ , which equals $-x/y$ , and the lower half checks the same way.

General and particular solutions

Definition A general solution of an equation on an interval $I$ is a formula with arbitrary constants that produces every solution on $I$ as the constants vary. For a linear equation of order $n$ whose coefficients and forcing term are continuous on $I$ , with a leading coefficient that never vanishes on $I$ , it contains exactly $n$ constants. A particular solution is one member of the family, with every constant fixed.

In a nonhomogeneous linear equation the phrase has a second use. There a particular solution $y_p$ is any one solution of the forced equation, chosen without regard to initial data, and the general solution is $y_p$ plus the general solution of the homogeneous equation with the same left side.

In plain words The general solution is the whole collection of answers, with its constants still free. A particular solution is one curve from that collection. The surrounding problem tells you whether a particular solution means the member selected by the data or just some solution of a forced equation.

Example Every solution of $y' = 2x$ on $\mathbb{R}$ has the form $x^2 + C$ , because two functions with the same derivative on an interval differ by a constant. So $y = x^2 + C$ is the general solution, and $y = x^2 + 3$ is a particular solution. For $y'' + y = 0$ the general solution is $y = C_1\cos x + C_2\sin x$ , with two constants for a second-order equation. For the forced equation $y'' + y = x$ , the function $y_p = x$ is a particular solution, since $y_p'' + y_p = 0 + x$ , and the general solution is $y = x + C_1\cos x + C_2\sin x$ .

A family found by a calculation can miss solutions. The family $y = 1/(C - x)$ solves $y' = y^2$ , and so does the zero function, which no choice of $C$ produces, so that family alone is not the general solution.

Solution verification

Definition To verify that a function $\varphi$ solves a problem on an interval $I$ , check three things separately. First, $\varphi$ is defined on $I$ and has there every derivative the equation needs. Second, substituting $\varphi$ and its derivatives into the equation gives an identity, true for every $x$ in $I$ . Third, $\varphi$ meets each initial or boundary condition. Passing one check says nothing about the others.

In plain words You never have to solve an equation to test an answer. Substitute and simplify, then check the data and the interval.

Example Test $y = x^2 + 3$ against $y' = 2x$ with $y(1) = 4$ . Its derivative is $2x$ for every real $x$ , and $y(1) = 1 + 3 = 4$ , so it passes every check on $\mathbb{R}$ . The function $y = x^2 + 5$ has the right derivative but gives $y(1) = 6$ . The function $y = 4x$ passes through $(1, 4)$ , yet its derivative $4$ equals $2x$ only at $x = 2$ , so it does not satisfy the equation on any interval. Each wrong candidate fails exactly one of the checks.

Domain of a solution

Definition The domain of a solution is the interval $I$ on which it is claimed to solve the equation. It must be an interval. It must leave out every point where the function or one of its needed derivatives fails to exist and every point where the equation itself is undefined. For an initial-value problem it must contain the initial point. The set where a formula makes sense can be larger than this interval, or not an interval at all.

In plain words A solution is a formula together with an interval. The same formula on two separate intervals counts as two different solutions.

Example The function $y = 1/x$ solves $y' = -y^2$ , since $y' = -1/x^2 = -y^2$ . Its formula makes sense for every $x \ne 0$ , which is not an interval, so it gives one solution on $(0, \infty)$ and another on $(-\infty, 0)$ . With the condition $y(1) = 1$ , the solution is $y = 1/x$ on $(0, \infty)$ only. In the same way, the upper half $y = \sqrt{25 - x^2}$ of the circle is continuous on $[-5, 5]$ , but at $x = \pm 5$ it has vertical tangents and the equation $y' = -x/y$ would divide by zero. Its domain as a solution is the open interval $(-5, 5)$ .

Initial-value problem

Definition An initial-value problem for an equation of order $n$ consists of the equation together with the values of $y, y', \ldots, y^{(n-1)}$ at one point $x_0$ . For a first-order equation it reads

y' = f(x, y), \qquad y(x_0) = y_0.

A solution is a function on an open interval containing $x_0$ that solves the equation there and meets every condition. For a linear equation of order $n$ whose coefficients and forcing term are continuous on an open interval $I$ , with a leading coefficient that never vanishes on $I$ , every initial-value problem with $x_0$ in $I$ has exactly one solution, and that solution exists on all of $I$ .

In plain words Initial data describe the state at one moment, a height for a first-order equation, and a height and a slope for a second-order one. For linear equations that meet these conditions, that is exactly enough information to pick out one solution.

Example Solve $y'' + y = 0$ with $y(0) = 2$ and $y'(0) = -1$ . Start from $y = C_1\cos x + C_2\sin x$ , so $y(0) = C_1 = 2$ . The derivative is $y' = -C_1\sin x + C_2\cos x$ , so $y'(0) = C_2 = -1$ . The solution is

y = 2\cos x - \sin x

on all of $\mathbb{R}$ . As a check, $y'' = -2\cos x + \sin x = -y$ , the value at $0$ is $2$ , and the slope at $0$ is $-1$ .

Boundary-value problem

Definition A boundary-value problem imposes conditions at two or more different points, usually the endpoints $a < b$ of an interval, for example $y(a) = \alpha$ and $y(b) = \beta$ for a second-order equation. Other boundary conditions prescribe $y'$ , or a combination of $y$ and $y'$ , at an endpoint. A solution satisfies the equation on the interval from $a$ to $b$ and every boundary condition. Even for a linear equation with continuous coefficients and forcing term and a leading coefficient that never vanishes, where every initial-value problem has exactly one solution, a boundary-value problem can have exactly one solution, infinitely many or none, and which case occurs is decided by solving the boundary equations for the constants.

In plain words The data are split between two ends. The equation fixes how the curve bends, and the question is whether that bending can be made to hit both targets.

Example Take $y'' + y = 0$ with general solution $y = C_1\cos x + C_2\sin x$ . The condition $y(0) = 0$ forces $C_1 = 0$ and leaves $y = C_2\sin x$ . Each of the three cases then occurs.

With $y(\pi/2) = 1$ on $[0, \pi/2]$ , the second condition gives $C_2 = 1$ , so the only solution is $y = \sin x$ .
With $y(\pi) = 0$ on $[0, \pi]$ , every $C_2$ works, because $\sin\pi = 0$ , so there are infinitely many solutions.
With $y(\pi) = 1$ on $[0, \pi]$ , the second condition reads $0 = 1$ for every $C_2$ , so there is no solution.

Check yourself

Find the solution of $y' = 3x^2$ with $y(2) = 5$ , and verify it.

Show the answer

Every solution on $\mathbb{R}$ has the form $y = x^3 + C$ . The condition gives $8 + C = 5$ , so $C = -3$ and $y = x^3 - 3$ . To verify it, note that the derivative of $x^3 - 3$ is $3x^2$ for every real $x$ and that $y(2) = 8 - 3 = 5$ . The solution is valid on all of $\mathbb{R}$ .

A classmate solves $y'' + y = 0$ with $y(0) = 1$ and $y(\pi) = -1$ , finds $y = \cos x$ , and calls it the solution. What is wrong with that conclusion? What changes if the second condition is $y(\pi) = 1$ instead?

Show the answer

Start from $y = C_1\cos x + C_2\sin x$ . The first condition gives $C_1 = 1$ . The second gives $y(\pi) = -C_1 = -1$ , which then holds automatically and puts no restriction on $C_2$ . So every function $y = \cos x + C_2\sin x$ solves the problem, and $\cos x$ is only the case $C_2 = 0$ . The boundary-value problem has infinitely many solutions.

With $y(\pi) = 1$ the second condition demands $-C_1 = 1$ , which contradicts $C_1 = 1$ , so there is no solution at all.

Show that the relation $y^2 = x$ is an implicit solution of $y' = 1/(2y)$ . Find the explicit solution through the point $(4, -2)$ and give its interval.

Show the answer

Differentiating $y^2 = x$ with respect to $x$ gives $2yy' = 1$ , so $y' = 1/(2y)$ wherever $y \ne 0$ . The point $(4, -2)$ lies on the lower branch, so the explicit solution is $y = -\sqrt{x}$ . Its derivative is

y' = -\frac{1}{2\sqrt{x}} = \frac{1}{2y},

since $2y = -2\sqrt{x}$ . The interval is $(0, \infty)$ . At $x = 0$ the branch reaches $y = 0$ , where the equation would divide by zero and $-\sqrt{x}$ is not differentiable, and for $x < 0$ no real $y$ satisfies $y^2 = x$ .

Solutions, Initial Conditions, and Boundary Conditions

Key terms

Explicit and implicit solutions

General and particular solutions

Solution verification

Domain of a solution

Initial-value problem

Boundary-value problem

Check yourself

How this lesson connects

Builds on

Used later in