ODE-03

First-Order Methods

Exact Equations and Integrating Factors

Learn to identify and solve exact differential equations, find integrating factors to convert non-exact equations, and understand first-order implicit equations including the elegant Clairaut equation.

Learning Objectives

Identify and solve exact differential equations
Apply the exactness criterion for differential forms
Master common total differential formulas
Find integrating factors depending on one variable
Use the grouped integrating factor method
Solve Clairaut equations and find singular solutions
Understand the geometric meaning of envelopes
Apply integrating factor techniques to applications

1. Exact Differential Equations

A first-order differential equation in the form $M(t,x)dt + N(t,x)dx = 0$ is called exactif the left side is the total differential of some function $u(t,x)$ . This means we can find the general solution by simply finding this potential function.

Definition 3.1: Exact Equation

A differential equation $M(t,x)dt + N(t,x)dx = 0$ is called an exact equation(or total differential equation) if there exists a continuously differentiable function $u(t,x)$ such that:

du(t,x) = M(t,x)dt + N(t,x)dx

Equivalently, $\frac{\partial u}{\partial t} = M(t,x)$ and $\frac{\partial u}{\partial x} = N(t,x)$ .

The key question is: how do we determine if an equation is exact, and how do we find the potential function $u$ ?

Theorem 3.1: Exactness Criterion

Let $M(t,x), N(t,x) \in C(D)$ where $D \subset \mathbb{R}^2$ is a simply connected domain, and suppose $\frac{\partial M}{\partial x}$ and $\frac{\partial N}{\partial t}$ are continuous in $D$ . Then the equation $M\\,dt + N\\,dx = 0$ is exact if and only if:

\frac{\partial M}{\partial x} = \frac{\partial N}{\partial t}

When this condition holds, the general integral is given by:

u(t,x) = \int_{t_0}^{t} M(s, x)\,ds + \int_{x_0}^{x} N(t_0, s)\,ds = C

Proof of Theorem 3.1:

Necessity: If the equation is exact, there exists $u(t,x)$ such that $\frac{\partial u}{\partial t} = M$ and $\frac{\partial u}{\partial x} = N$ . Since $u$ has continuous second partial derivatives:

\frac{\partial M}{\partial x} = \frac{\partial^2 u}{\partial x \partial t} = \frac{\partial^2 u}{\partial t \partial x} = \frac{\partial N}{\partial t}

Sufficiency: Assume $\frac{\partial M}{\partial x} = \frac{\partial N}{\partial t}$ . We construct $u(t,x)$ as follows. Let:

u(t,x) = \int_{t_0}^{t} M(s, x)\,ds + \varphi(x)

where $\varphi(x)$ is a function to be determined. Then $\frac{\partial u}{\partial t} = M(t,x)$ . For $\frac{\partial u}{\partial x} = N$ :

\frac{\partial u}{\partial x} = \int_{t_0}^{t} \frac{\partial M}{\partial x}(s,x)\,ds + \varphi'(x) = \int_{t_0}^{t} \frac{\partial N}{\partial t}(s,x)\,ds + \varphi'(x)

= N(t,x) - N(t_0,x) + \varphi'(x)

Setting $\varphi'(x) = N(t_0,x)$ , we get $\frac{\partial u}{\partial x} = N(t,x)$ . ∎

Example 3.1: Solving an Exact Equation

Problem: Solve $(2t\sin x + 3t^2 x + t^4)dt + (t^3 + t^2\cos x + x^2)dx = 0$ .

Solution:

First, verify exactness. Let $M = 2t\sin x + 3t^2 x + t^4$ and $N = t^3 + t^2\cos x + x^2$ .

\frac{\partial M}{\partial x} = 2t\cos x + 3t^2, \quad \frac{\partial N}{\partial t} = 3t^2 + 2t\cos x

Since $\frac{\partial M}{\partial x} = \frac{\partial N}{\partial t}$ , the equation is exact.

Taking $(t_0, x_0) = (0, 0)$ , we compute:

u(t,x) = \int_0^t (2s\sin x + 3s^2 x + s^4)\,ds + \int_0^x s^2\,ds

= t^2\sin x + t^3 x + \frac{t^5}{5} + \frac{x^3}{3}

The general integral is:

t^2\sin x + t^3 x + \frac{t^5}{5} + \frac{x^3}{3} = C

2. Common Total Differential Formulas

Recognizing common total differentials can greatly simplify solving exact equations. Here are the most frequently used formulas:

Essential Total Differential Formulas

t\,dt + x\,dx = \frac{1}{2}d(t^2 + x^2)

\frac{x\,dt + t\,dx}{tx} = d(\ln(tx))

\frac{x\,dt - t\,dx}{t^2 + x^2} = d\left(\arctan\frac{t}{x}\right)

\frac{x\,dt - t\,dx}{x^2} = d\left(\frac{t}{x}\right)

\frac{x\,dt - t\,dx}{-t^2} = d\left(\frac{x}{t}\right)

\frac{t\,dt + x\,dx}{t^2 + x^2} = \frac{1}{2}d\ln(t^2 + x^2)

\frac{x\,dt - t\,dx}{t^2 - x^2} = \frac{1}{2}d\ln\left|\frac{t-x}{t+x}\right|

Remark:

These formulas are particularly useful when you can identify parts of a differential equation that match these patterns. Often, the key to solving an equation is rewriting it to reveal these standard forms.

3. Integrating Factors

Most first-order equations are not exact. However, we can often multiply by a suitable function called anintegrating factor to make the equation exact.

Definition 3.2: Integrating Factor

For the equation $M(t,x)dt + N(t,x)dx = 0$ , a non-zero function $\mu(t,x) \in C(D)$ is called an integrating factor if the equation:

\mu(t,x)M(t,x)dt + \mu(t,x)N(t,x)dx = 0

is an exact equation.

Theorem 3.2: Condition for Integrating Factor

A function $\mu(t,x)$ is an integrating factor for $M\\,dt + N\\,dx = 0$ if and only if:

N\frac{\partial \mu}{\partial t} - M\frac{\partial \mu}{\partial x} = \mu\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right)

This PDE is generally harder to solve than the original equation. However, in special cases, we can find integrating factors that depend on only one variable.

Theorem 3.3: Integrating Factor Depending on One Variable

Case 1: If $\frac{1}{N}\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right) = G(t)$ depends only on $t$ , then:

\mu(t) = e^{\int G(t)\,dt}

is an integrating factor.

Case 2: If $-\frac{1}{M}\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right) = H(x)$ depends only on $x$ , then:

\mu(x) = e^{\int H(x)\,dx}

is an integrating factor.

Proof of Theorem 3.3:

We prove Case 2. Suppose $\mu = \mu(x)$ depends only on $x$ . Then $\frac{\partial \mu}{\partial t} = 0$ , and the condition becomes:

-M\frac{d\mu}{dx} = \mu\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right)

Rearranging:

\frac{1}{\mu}\frac{d\mu}{dx} = -\frac{1}{M}\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right) = H(x)

Integrating: $\ln|\mu| = \int H(x)\,dx$ , giving $\mu(x) = e^{\int H(x)\,dx}$ . ∎

Algorithm 3.1: Finding an Integrating Factor

Input: Equation $M(t,x)dt + N(t,x)dx = 0$

Output: An integrating factor $\mu$ if one exists in simple form

1. Check if equation is already exact: $\frac{\partial M}{\partial x} \stackrel{?}{=} \frac{\partial N}{\partial t}$

2. Compute $R = \frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}$

3. If $R/N$ depends only on $t$ : $\mu(t) = e^{\int (R/N)\,dt}$

4. Else if $-R/M$ depends only on $x$ : $\mu(x) = e^{\int (-R/M)\,dx}$

5. Else try special forms: $\mu(tx), \mu(t^2+x^2), \mu(t/x)$ , etc.

6. return Integrating factor $\mu$

Example 3.2: Finding an Integrating Factor

Problem: Solve $x\\,dt + (t^2 + t + x^2)dx = 0$ .

Solution:

Here $M = x$ and $N = t^2 + t + x^2$ . Check exactness:

\frac{\partial M}{\partial x} = 1, \quad \frac{\partial N}{\partial t} = 2t + 1

Not exact. Let's make the substitution $y = t + 1$ (so $dy = dt$ ):

x\,dy + (y^2 - y + x^2)dx = 0

Rewrite as:

x\,dy - y\,dx + (x^2 + y^2)dx = 0

Using the integrating factor $\mu = \frac{1}{x^2 + y^2}$ :

\frac{x\,dy - y\,dx}{x^2 + y^2} + dx = 0

This gives $d\left(\arctan\frac{y}{x}\right) + dx = 0$ , so:

\arctan\frac{y}{x} + x = C \implies \arctan\frac{t+1}{x} + x = C

4. Grouped Integrating Factor Method

For equations that can be split into several groups, each with its own integrating factor, we can use a powerful technique to find a common integrating factor.

Theorem 3.4: Grouped Integrating Factor Theorem

Suppose the non-exact equation $\sum_{i=1}^{n}(M_i\,dt + N_i\,dx) = 0$ can be grouped such that each group $M_i\\,dt + N_i\\,dx = 0$ has integrating factor $\mu_i$ satisfying:

\mu_i(M_i\,dt + N_i\,dx) = d\Phi_i(t,x)

Then for any differentiable function $g$ , the function $\mu_i g(\Phi_i)$ is also an integrating factor for the $i$ -th group equation.

Proof of Theorem 3.4:

For any $i = 1, \ldots, n$ , we compute:

\frac{\partial}{\partial x}(\mu_i g(\Phi_i)M_i) = \frac{\partial}{\partial x}(\mu_i M_i)g(\Phi_i) + \frac{\partial \Phi_i}{\partial x}g'(\Phi_i)\mu_i M_i

Since $\mu_i M_i = \frac{\partial \Phi_i}{\partial t}$ and $\mu_i N_i = \frac{\partial \Phi_i}{\partial x}$ :

= \frac{\partial}{\partial t}(\mu_i N_i)g(\Phi_i) + \mu_i^2 M_i N_i g'(\Phi_i)

= \frac{\partial}{\partial t}(\mu_i N_i)g(\Phi_i) + \frac{\partial \Phi_i}{\partial t}g'(\Phi_i)\mu_i N_i = \frac{\partial}{\partial t}(\mu_i g(\Phi_i)N_i)

Thus $\mu_i g(\Phi_i)$ is an integrating factor. ∎

Algorithm 3.2: Grouped Integrating Factor Method

Input: Equation that can be split as $\sum (M_i\,dt + N_i\,dx) = 0$

Output: General integral

1. Split the equation into groups

2. For each group, find $\mu_i$ and $\Phi_i$

3. Choose functions $g_i$ such that $\mu_1 g_1(\Phi_1) = \mu_2 g_2(\Phi_2) = \cdots = \mu$

4. Multiply original equation by common $\mu$

5. Integrate to find general integral

Example 3.3: Grouped Integrating Factor Method

Problem: Solve $\left(\frac{x}{t} + 3t^2\right)dt + \left(1 + \frac{t^3}{x}\right)dx = 0$ .

Solution:

Group the equation as:

\left(\frac{x}{t}\,dt + dx\right) + \left(3t^2\,dt + \frac{t^3}{x}\,dx\right) = 0

Group 1: $\frac{x}{t}\,dt + dx = 0$

Integrating factor: $\mu_1 = t$ , giving $x\,dt + t\,dx = d(tx)$ , so $\Phi_1 = tx$ .

Group 2: $3t^2\,dt + \frac{t^3}{x}\,dx = 0$

Integrating factor: $\mu_2 = x$ , giving $3t^2 x\,dt + t^3\,dx = d(t^3 x)$ , so $\Phi_2 = t^3 x$ .

Find $g_1, g_2$ such that $tg_1(tx) = xg_2(t^3 x)$ .

Take $g_1(s) = s^2$ and $g_2(s) = s$ , giving $\mu = t(tx)^2 = t^3 x^2$ .

Multiply original equation by $\mu = t^3 x^2$ :

(x^3 t^2\,dt + t^3 x^2\,dx) + (3t^5 x^2\,dt + t^6 x\,dx) = 0

= d\left(\frac{1}{3}x^3 t^3\right) + d\left(\frac{1}{2}t^6 x^2\right) = 0

The general integral is:

\frac{1}{3}x^3 t^3 + \frac{1}{2}t^6 x^2 = C

5. First-Order Implicit Equations and Clairaut Equations

When the derivative $x'$ cannot be explicitly solved, we have an implicit equation $F(t, x, x') = 0$ . A particularly elegant case is the Clairaut equation.

Definition 3.3: Clairaut Equation

A Clairaut equation has the form:

x = tx' + f(x')

where $f$ is a given function with $f'' \not\equiv 0$ . Setting $p = x'$ , the equation becomes $x = tp + f(p)$ .

Theorem 3.5: Solution of Clairaut Equations

The Clairaut equation $x = tx' + f(x')$ has:

General solution: $x = Ct + f(C)$ — a family of straight lines
Singular solution: Given parametrically by $t = -f'(p)$ , $x = -pf'(p) + f(p)$

Proof of Theorem 3.5:

Let $p = x'$ . Differentiate $x = tp + f(p)$ with respect to $t$ :

p = p + t\frac{dp}{dt} + f'(p)\frac{dp}{dt} = p + (t + f'(p))\frac{dp}{dt}

This gives:

(t + f'(p))\frac{dp}{dt} = 0

Case 1: $\frac{dp}{dt} = 0$ , so $p = C$ (constant). Substituting back: $x = Ct + f(C)$ .

Case 2: $t + f'(p) = 0$ , giving $t = -f'(p)$ . Substituting into the original equation:

x = -pf'(p) + f(p)

This parametric form defines the singular solution. ∎

Definition 3.4: Singular Solution

A singular solution of a differential equation is a solution that cannot be obtained from the general solution by any choice of the arbitrary constant, and at every point on its curve, it is tangent to some member of the family of curves representing the general solution.

Geometrically, the singular solution is the envelope of the family of general solution curves.

Example 3.4: Solving a Clairaut Equation

Problem: Solve $x = tx' - \frac{1}{4}(x')^2$ .

Solution:

This is a Clairaut equation with $f(p) = -\frac{1}{4}p^2$ .

General solution: $x = Ct - \frac{1}{4}C^2$ (a family of lines).

Singular solution: Since $f'(p) = -\frac{1}{2}p$ , we have:

t = -f'(p) = \frac{1}{2}p \implies p = 2t

Substituting:

x = t(2t) - \frac{1}{4}(2t)^2 = 2t^2 - t^2 = t^2

The singular solution is $x = t^2$ (a parabola).

Geometric interpretation: The parabola $x = t^2$ is the envelope of the family of lines $x = Ct - \frac{C^2}{4}$ . Each line is tangent to the parabola.

Remark:

The singular solution often has physical significance. For example, in optics, if each line represents a light ray, the envelope (singular solution) represents a caustic curve where light concentrates.

6. Applications

Example 3.5: Searchlight Mirror Shape

Problem: Find the shape of a mirror such that light rays from a point source at the origin reflect as parallel rays.

Solution:

Let point $P(t, x)$ be on the mirror curve. By the law of reflection, if $\alpha$ is the angle between the tangent and the reflected ray (parallel to $t$ -axis), then:

x'(t) = \tan\alpha

Since the angle of incidence equals the angle of reflection, $|OA| = |OP|$ where $A$ is on the $t$ -axis. This gives:

\frac{x}{x'} - t = \sqrt{t^2 + x^2}

Let $u = t/x$ . After substitution and integration:

x = 2Ct + \frac{C}{2}

This is a parabola, confirming that parabolic mirrors produce parallel reflected rays from a point source at the focus.

Practice Quiz

Exact Equations and Integrating Factors Quiz

Questions

Correct

Accuracy

For the equation

M(t,x)dt + N(t,x)dx = 0

to be exact, which condition must hold?

Easy

Not attempted

Which of the following is a common total differential formula?

Easy

Not attempted

What is an integrating factor for the equation

M\,dt + N\,dx = 0

Easy

Not attempted

Is the equation

(2t\sin x + 3t^2x)dt + (t^2\cos x + t^3)dx = 0

exact?

Medium

Not attempted

\frac{1}{M}\left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial t}\right)

depends only on

x

, then the integrating factor

\mu

depends on:

Medium

Not attempted

The general form of a Clairaut equation is:

Medium

Not attempted

What type of solutions does a Clairaut equation have?

Medium

Not attempted

For the equation

x\,dt + (t^2 + t + x^2)dx = 0

, which substitution helps find an integrating factor?

Medium

Not attempted

In the grouped integrating factor method, if equation

M_i\,dt + N_i\,dx = 0

has integrating factor

\mu_i

with first integral

\Phi_i

, then

\mu_i g(\Phi_i)

is also an integrating factor for any:

Hard

Not attempted

The singular solution of a Clairaut equation is geometrically the:

Hard

Not attempted

Frequently Asked Questions

How do I know if an equation is exact?

Check if

\frac{\partial M}{\partial x} = \frac{\partial N}{\partial t}

. If this equality holds and the domain is simply connected, the equation

M\\,dt + N\\,dx = 0

is exact. Remember that

M

is the coefficient of

dt

and

N

is the coefficient of

dx

What if I can't find an integrating factor?

Not every equation has a simple integrating factor. Try these approaches in order: (1) Check if

R/N

depends only on

t

, (2) Check if

-R/M

depends only on

x

, (3) Try special forms like

\mu(tx)

\mu(t^2+x^2)

, (4) Try the grouped integrating factor method. If none work, the equation may require a different solution technique.

What is the geometric meaning of the singular solution?

The singular solution is the envelope of the family of curves that form the general solution. Geometrically, at each point of the singular solution curve, exactly one member of the general solution family is tangent to it. This is why the singular solution cannot be obtained by choosing a specific constant in the general solution.

How do I verify my solution to an exact equation?

After finding

u(t,x) = C

, verify by checking: (1)

\frac{\partial u}{\partial t} = M

and (2)

\frac{\partial u}{\partial x} = N

. Also, you can differentiate

u(t,x) = C

implicitly to recover the original differential equation.

When should I use the grouped integrating factor method?

Use this method when: (1) The equation naturally splits into groups, (2) Each group has a recognizable integrating factor, (3) Standard integrating factor methods don't work. The key is finding functions

g_i

such that all

\mu_i g_i(\Phi_i)

are equal.

Can a Clairaut equation have no singular solution?

Yes, if

f''(p) \equiv 0

, meaning

f(p)

is linear in

p

. In this case, the equation reduces to

x = tp + ap + b = (t+a)p + b

, which is a family of lines through a single point, and there is no envelope (singular solution).