Differential Equations And Initial Value Problems

Many problems in mathematics, science, engineering, and economics are posed in terms of an equation involving derivatives of an unknown function. An equation that involves derivatives of an unknown function is called a differential equation.

Term	Meaning
Differential equation	An equation involving derivatives of an unknown function
General solution	The family $F (x) + C$ satisfying the differential equation
Initial condition	The value of the unknown function at a specific point, e.g., $y (x_{0}) = y_{0}$
Particular solution	The unique solution satisfying both the equation and the initial condition
Initial-value problem	A differential equation together with an initial condition
Separation of variables	A technique: rearrange so all $y$ -terms are on one side and all $x$ -terms on the other, then integrate both sides

Differential Equations

For example, f'(x) = 3x^2 - 2x or f'(x) = -x/f(x) are two differential equations. They are often written as

\begin{aligned} y' &= 3x^2 - 2x, \\ \text{and}\quad y' &= -\frac{x}{y}. \end{aligned}

Notice the difference between the two. In the first, the right-hand side involves only $x$ , so the unknown function can be recovered by integrating directly. In the second, the unknown $y$ appears on the right as well, and a little more work is needed.

General and Particular Solutions

We know that every function that satisfies f'(x) = 3x^2 - 2x (or y' = 3x^2 - 2x) is of the form

y = f (x) = x^{3} - x^{2} + C .

This is called the general solution of the given differential equation.

The general solution of a differential equation involves an arbitrary constant (or constants).

If we know the value of $f$ for one value of $x$ , then we can determine a particular solution. For example, if we know $f (1) = 2$ , then we obtain

f (x) = x^{3} - x^{2} + 2.

This is called a particular solution and the extra information $f (1) = 2$ is called an initial condition.

If this differential equation is written as y' = 3x^2 - 2x, it is common practice to write the given initial condition as $y (1) = 2$ . The initial condition $f (1) = 2$ specifies only one curve in the family of curves that passes through the point $(1, 2)$ .

Initial-Value Problems

The problem of solving a differential equation subject to an initial condition is called an initial-value problem.

Therefore

y' = 3x^2 - 2x, \quad y(1) = 2

is an initial-value problem.

An initial-value problem always has two parts, and both are needed: the differential equation narrows the answer down to a family of curves, and the initial condition selects one member of that family.

Separation of Variables

Recall that y' can be regarded as $d y$ divided by $d x$ . Therefore, the problem y' = 3x^2 - 2x is equivalent to

\frac{d y}{d x} = 3 x^{2} - 2 x

d y = (3 x^{2} - 2 x) d x .

Now we can integrate each side:

\int d y = \int (3 x^{2} - 2 x) d x

\Rightarrow y + C_{1} = x^{3} - x^{2} + C_{2} .

We can combine $C_{1}$ and $C_{2}$ and simply write

y = x^{3} - x^{2} + C,

where $C = C_{2} - C_{1}$ . This technique, called separation of variables, can help us solve some differential equations.

The Method Step by Step

Separation of Variables

Separate. Rewrite the equation so that every $y$ sits with $d y$ on one side and every $x$ sits with $d x$ on the other.
Integrate each side with respect to its own variable, and combine the two constants of integration into a single constant $C$ .
Apply the initial condition to determine $C$ , which turns the general solution into the particular solution.

In words: because each side of the separated equation contains only one variable, each integral becomes an ordinary one-variable integral of the kind studied in the previous sections. The method only works when the equation can actually be split this way.

Worked Example: Finding a Curve from Its Slope

Determine the equation of a curve such that the slope of the tangent to the curve at any point is the negative ratio of the $x$ -value to the $y$ -value (the abscissa to the ordinate) and it passes through the point $(3, 4)$ .

Solution

This is an initial-value problem which can be mathematically expressed as

\frac{d y}{d x} = - \frac{x}{y}, y (3) = 4.

Step 1: separate the variables. We can solve the differential equation $d y / d x = - x / y$ by separation of variables. That is, we multiply both sides by $y d x$ , obtaining

y d y = - x d x .

Step 2: integrate each side. Integrating each side gives us

\frac{1}{2} y^{2} + C_{1} = - \frac{1}{2} x^{2} + C_{2}

x^2 + y^2 = C, \tag{i}

where $C = 2 (C_{2} - C_{1})$ . Equation (i) is the general solution of y' = -x/y. Several curves of the form $x^{2} + y^{2} = C$ are shown in the figure below.

Step 3: apply the initial condition. If we impose the initial condition that the curve must pass through the point $(3, 4)$ or $y (3) = 4$ , we obtain

\underset{3^{2}}{\underset{⏟}{x^{2}}} + \underset{4^{2}}{\underset{⏟}{y^{2}}} = C \Rightarrow C = 25.

Hence, the particular solution satisfying the differential equation and the initial condition is the equation of the circle $x^{2} + y^{2} = 25$ .

Concentric circles x² + y² = C for various values of C, with the circle x² + y² = 25 passing through the point (3,4) highlighted. — The curves $x^{2} + y^{2} = C$ for various values of $C$ . We identify the curve $x^{2} + y^{2} = 25$ as the one that passes through the point $(3, 4)$ .

Exercises

Verify that each equation is a general solution of the differential equation beside it:

(a) $y^{2} = x^{2} + C$ for y\,y' = x

(b) $y = C x$ for x\,y' = y

(c) Which member of the family in part (b) passes through $(2, - 6)$ ?

Answer

(a) Differentiating implicitly gives 2y\,y' = 2x, that is y\,y' = x.

(b) y' = C, so x\,y' = Cx = y.

(c) $y = - 3 x$

Solution

To check that a family is a general solution, differentiate it and confirm that the arbitrary constant disappears, leaving exactly the given differential equation.

(a) Differentiate

y^{2} = x^{2} + C

with respect to

x

, treating

y

as a function of

x

: 2y\,\frac{dy}{dx} = 2x \quad\Longrightarrow\quad y\,y' = x.

The constant $C$ vanishes on differentiation, which is why a single family covers every solution.

(b) From

y = C x

we get y' = C. Substituting into the left side, x\,y' = x\cdot C = Cx = y,

so the equation is satisfied for every value of $C$ . Geometrically this family is all straight lines through the origin.

(c) Substituting

x = 2

y = - 6

into

y = C x

: \begin{aligned} -6 &= 2C \\ C &= -3, \end{aligned}

so the particular solution is $y = - 3 x$ .

Solve each initial-value problem by direct integration:

(a) y' = 6x^{2}-4, $y (1) = 2$

(b) $\frac{d y}{d x} = \sec^{2} x$ , $y (0) = 3$

(c) $\frac{d y}{d x} = \sqrt{x} + \frac{1}{\sqrt{x}}$ , $y (1) = 0$

Answer

(a) $y = 2 x^{3} - 4 x + 4$

(b) $y = \tan x + 3$

(c) $y = \frac{2}{3} x^{3 / 2} + 2 \sqrt{x} - \frac{8}{3}$

Solution

When the right-hand side involves $x$ alone, no separation is needed: integrate directly to get the general solution, then use the initial condition to pick out the particular solution.

(a) Integrating,

y = 2 x^{3} - 4 x + C .

Imposing $y (1) = 2$ :

\begin{aligned} 2-4+C &= 2 \\ C &= 4, \end{aligned}

so $y = 2 x^{3} - 4 x + 4$ .

(b) Since

\int \sec^{2} x d x = \tan x + C

y = \tan x + C .

Because $\tan 0 = 0$ , the condition $y (0) = 3$ gives $C = 3$ , so $y = \tan x + 3$ .

(c) Write the integrand as powers:

\sqrt{x} + \frac{1}{\sqrt{x}} = x^{1 / 2} + x^{- 1 / 2}

. Then \begin{aligned} y &= \frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+C \\ &= \frac{2}{3}x^{3/2}+2\sqrt{x}+C. \end{aligned}

Imposing $y (1) = 0$ :

\begin{aligned} \frac{2}{3}+2+C &= 0 \\ C &= -\frac{8}{3}, \end{aligned}

so $y = \frac{2}{3} x^{3 / 2} + 2 \sqrt{x} - \frac{8}{3}$ .

Solve the initial-value problem

\frac{d y}{d x} = \frac{x^{2}}{y^{2}}, y (0) = 1,

by separation of variables.

Answer

$y^{3} = x^{3} + 1$ , that is $y = \sqrt[3]{x^{3} + 1}$ .

Solution

Step 1: separate the variables. Multiply both sides by $y^{2} d x$ so that every $y$ sits with $d y$ and every $x$ with $d x$ :

y^{2} d y = x^{2} d x .

Step 2: integrate each side. \begin{aligned} \int y^{2}\,dy &= \int x^{2}\,dx \\ \frac{y^{3}}{3}+C_1 &= \frac{x^{3}}{3}+C_2. \end{aligned}

Multiplying by $3$ and absorbing the two constants into one, $C = 3 (C_{2} - C_{1})$ , gives the general solution

y^{3} = x^{3} + C .

Step 3: apply the initial condition. Substituting

x = 0

y = 1

1 = 0 + C ⟹ C = 1.

Hence $y^{3} = x^{3} + 1$ , or explicitly $y = \sqrt[3]{x^{3} + 1}$ .

Check. Differentiating $y^{3} = x^{3} + 1$ implicitly gives 3y^{2}y' = 3x^{2}, so y' = x^{2}/y^{2} as required, and $y (0) = \sqrt[3]{1} = 1$ .

Solve the initial-value problem

\frac{d y}{d x} = 3 x^{2} y^{2}, y (0) = 1.

Answer

$y = \frac{1}{1 - x^{3}}$

Solution

Step 1: separate. Divide by $y^{2}$ and multiply by $d x$ :

\frac{d y}{y^{2}} = 3 x^{2} d x .

Step 2: integrate. The left side is a power with exponent

- 2

: \begin{aligned} \int y^{-2}\,dy &= \int 3x^{2}\,dx \\ -\frac{1}{y} &= x^{3}+C. \end{aligned}
Step 3: apply the initial condition. Substituting

x = 0

y = 1

- 1 = 0 + C ⟹ C = - 1.

Therefore $- \frac{1}{y} = x^{3} - 1$ , and solving for $y$ ,

y = \frac{- 1}{x^{3} - 1} = \frac{1}{1 - x^{3}} .

Notice that this solution blows up as $x \to 1$ . A perfectly ordinary differential equation can have a solution that exists only on a limited interval, here $- \infty < x < 1$ , the interval containing the initial point $x = 0$ .

Solve the initial-value problem

\frac{d y}{d x} = \frac{y}{x}, y (1) = - 3, x > 0.

Answer

$y = - 3 x$

Solution

Step 1: separate. Here both variables appear, so send each to its own side:

\frac{d y}{y} = \frac{d x}{x} .

Step 2: integrate. Both sides are of the logarithmic form

\int \frac{d u}{u} = \ln | u | + C

\ln | y | = \ln | x | + C_{1} .

Exponentiating, and writing $K = e^{C_{1}} > 0$ ,

| y | = K | x | ⟹ y = \pm K x .

Allowing the constant to take either sign, the general solution is $y = C x$ , the same family verified in Exercise 1(b).

Step 3: apply the initial condition. From

y (1) = - 3

- 3 = C (1) ⟹ C = - 3,

so $y = - 3 x$ .

Check. If $y = - 3 x$ then y' = -3, while $\frac{y}{x} = \frac{- 3 x}{x} = - 3$ . The two agree.

Determine the equation of the curve whose slope at any point is the negative ratio of the ordinate to the abscissa, and which passes through the point $(2, 3)$ . Compare the family you obtain with the family of circles found in the worked example of this section.

Answer

$x y = 6$ , one member of the family of rectangular hyperbolas $x y = C$ .

Solution

The ordinate is $y$ and the abscissa is $x$ , so "the negative ratio of the ordinate to the abscissa" is $- \frac{y}{x}$ . The initial-value problem is

\frac{d y}{d x} = - \frac{y}{x}, y (2) = 3.

Step 1: separate and integrate. \begin{aligned} \frac{dy}{y} &= -\frac{dx}{x} \\ \ln|y| &= -\ln|x|+C_1 \\ \ln|y|+\ln|x| &= C_1. \end{aligned}

Using $\ln a + \ln b = \ln (a b)$ , this is $\ln | x y | = C_{1}$ , so

x y = C .

Step 2: apply the initial condition. Substituting

(2, 3)

C = (2) (3) = 6,

so the required curve is $x y = 6$ , or $y = \frac{6}{x}$ .

Comparison. In the worked example the slope was $- \frac{x}{y}$ and the solutions were the circles $x^{2} + y^{2} = C$ . Here the slope is $- \frac{y}{x}$ and the solutions are the rectangular hyperbolas $x y = C$ . Exchanging the roles of $x$ and $y$ in the differential equation changes the family completely, which is a useful reminder that in $\frac{d y}{d x} = - \frac{x}{y}$ the placement of each variable matters.

Two questions about the worked example, $\frac{d y}{d x} = - \frac{x}{y}$ with $y (3) = 4$ :

(a) A student integrates both sides with respect to $x$ and writes $y = - \frac{x^{2}}{2} + C$ . What went wrong?

(b) In the general solution $x^{2} + y^{2} = C$ , which values of $C$ give an actual curve? Given the initial condition, on what portion of the circle $x^{2} + y^{2} = 25$ does the solution actually live?

Answer

(a) The right side contains the unknown $y$ , so $\int - \frac{x}{y} d x$ cannot be evaluated as if $y$ were constant. The variables must be separated first.

(b) Only $C > 0$ gives a circle; $C = 0$ gives the single point at the origin and $C < 0$ gives nothing real. The solution is the upper semicircle $y = \sqrt{25 - x^{2}}$ on $- 5 < x < 5$ .

Solution

(a) Writing

y = \int - \frac{x}{y} d x

treats

y

as a constant while integrating, but

y

is precisely the unknown function of

x

that we are looking for. Pulling it out of the integral assumes the answer. This is exactly the situation that separation of variables is designed for: multiply through by

y d x

first,

y d y = - x d x,

so that each side contains only one variable and each integral is a genuine one-variable integral. The correct general solution is $x^{2} + y^{2} = C$ , not $y = - \frac{x^{2}}{2} + C$ .

(b) Since

x^{2} + y^{2} \geq 0

for all real

x

and

y

, the equation

x^{2} + y^{2} = C

has real points only when

C \geq 0

. For

C > 0

it is a circle of radius

\sqrt{C}

centered at the origin, for

C = 0

it collapses to the single point

(0, 0)

, and for

C < 0

there are no points at all. So only

C > 0

produces a curve.

The initial condition $y (3) = 4$ gives $C = 25$ , the circle of radius $5$ . But a solution of a differential equation must be a function, and the full circle fails the vertical line test. Moreover $\frac{d y}{d x} = - \frac{x}{y}$ is undefined where $y = 0$ , which rules out the two endpoints $(\pm 5, 0)$ . Since the initial point $(3, 4)$ has $y > 0$ , the solution is the upper semicircle

y = \sqrt{25 - x^{2}}, - 5 < x < 5.

Differentiating confirms it: y' = \dfrac{-x}{\sqrt{25-x^{2}}} = -\dfrac{x}{y}.