Exact equations — exactness condition, finding potential function: Exercises

Throughout, an equation is written $M (x, y) d x + N (x, y) d y = 0$ , where:

$M$ is the ==coefficient multiplying $d x$ == (the " $x$ -slope" of the hidden hill),
$N$ is the ==coefficient multiplying $d y$ == (the " $y$ -slope"),
$M_{y}$ means $\frac{\partial M}{\partial y}$ (differentiate $M$ in $y$ , treating $x$ as frozen),
$N_{x}$ means $\frac{\partial N}{\partial x}$ (differentiate $N$ in $x$ , treating $y$ as frozen).

The picture to keep in your head:

Figure — Exact equations — exactness condition, finding potential function

Level 1 — Recognition

Goal: run the test correctly. No solving yet.

Exercise 1.1

Is $(3 x^{2} + 2 y) d x + (2 x - 5 y^{4}) d y = 0$ exact?

Recall Solution 1.1

Read off $M = 3 x^{2} + 2 y$ and $N = 2 x - 5 y^{4}$ .

$M_{y} = \frac{\partial}{\partial y} (3 x^{2} + 2 y) = 2$ (the $3 x^{2}$ has no $y$ , so it vanishes).
$N_{x} = \frac{\partial}{\partial x} (2 x - 5 y^{4}) = 2$ (the $- 5 y^{4}$ has no $x$ , so it vanishes).

$M_{y} = N_{x} = 2$ ⟹ exact. ✓

Exercise 1.2

Is $sin y d x + cos x d y = 0$ exact?

Recall Solution 1.2

$M = sin y$ , $N = cos x$ .

$M_{y} = cos y$ .
$N_{x} = - sin x$ .

$cos y \neq = - sin x$ in general ⟹ not exact. ✗

Exercise 1.3

Is $(y e^{x y}) d x + (x e^{x y}) d y = 0$ exact?

Recall Solution 1.3

$M = y e^{x y}$ , $N = x e^{x y}$ . These need the product rule.

$M_{y} = \frac{\partial}{\partial y} (y e^{x y}) = e^{x y} + y \cdot (x e^{x y}) = e^{x y} (1 + x y)$ .
$N_{x} = \frac{\partial}{\partial x} (x e^{x y}) = e^{x y} + x \cdot (y e^{x y}) = e^{x y} (1 + x y)$ .

Equal ⟹ exact. ✓ (This is $d (e^{x y})$ .)

Level 2 — Application

Goal: run the full I-D-M-I recipe on clean equations.

Exercise 2.1

Solve $(2 x + y) d x + (x + 2 y) d y = 0$ .

Recall Solution 2.1

$M = 2 x + y$ , $N = x + 2 y$ . Test: $M_{y} = 1$ , $N_{x} = 1$ ⟹ exact. Integrate $M$ in $x$ (treat $y$ as a constant): $F = \int (2 x + y) d x + g (y) = x^{2} + x y + g (y) .$ The $g (y)$ is the "constant" of $x$ -integration — it can depend on $y$ because $y$ was frozen. Differentiate in $y$ , match $N$ : $F_{y} = x + g^{'} (y) =! x + 2 y \Rightarrow g^{'} (y) = 2 y .$ Integrate: $g (y) = y^{2}$ . Answer: $x^{2} + x y + y^{2} = c .$

Exercise 2.2

Solve $(3 x^{2} y + 2) d x + (x^{3} + 4 y) d y = 0$ .

Recall Solution 2.2

$M = 3 x^{2} y + 2$ , $N = x^{3} + 4 y$ . Test: $M_{y} = 3 x^{2}$ , $N_{x} = 3 x^{2}$ ⟹ exact. Integrate $M$ : $F = \int (3 x^{2} y + 2) d x + g (y) = x^{3} y + 2 x + g (y) .$ Match $N$ : $F_{y} = x^{3} + g^{'} (y) = x^{3} + 4 y \Rightarrow g^{'} (y) = 4 y \Rightarrow g (y) = 2 y^{2} .$ Answer: $x^{3} y + 2 x + 2 y^{2} = c .$

Exercise 2.3

Solve $(cos y) d x - (x sin y - y^{2}) d y = 0$ .

Recall Solution 2.3

Be careful with the minus sign: $N = - (x sin y - y^{2}) = - x sin y + y^{2}$ . $M = cos y$ . Test: $M_{y} = - sin y$ ; $N_{x} = - sin y$ ⟹ exact. Integrate $M$ : $F = \int cos y d x + g (y) = x cos y + g (y)$ (here $cos y$ is a constant w.r.t. $x$ ). Match $N$ : $F_{y} = - x sin y + g^{'} (y) = - x sin y + y^{2} \Rightarrow g^{'} (y) = y^{2} \Rightarrow g (y) = \frac{1}{3} y^{3} .$ Answer: $x cos y + \frac{1}{3} y^{3} = c .$

Level 3 — Analysis

Goal: decide whether to use exactness, and diagnose failures.

Exercise 3.1

Show $y d x - x d y = 0$ is not exact, then note what fixes it.

Recall Solution 3.1

$M = y$ , $N = - x$ . $M_{y} = 1$ , $N_{x} = - 1$ . $1 \neq = - 1$ ⟹ not exact. What fixes it: multiply by the integrating factor $μ = 1/ x^{2}$ (a separate topic — see Integrating Factors for ODEs). Then $\frac{y}{x ^{2}} d x - \frac{1}{x} d y = 0, M = \frac{y}{x ^{2}}, N = - \frac{1}{x} .$ Now $M_{y} = 1/ x^{2}$ and $N_{x} = 1/ x^{2}$ ⟹ exact, and the left side is exactly $- d (y / x)$ . Solution: $y / x = c$ .

Exercise 3.2

For which constant $a$ is $(a x y + y^{2}) d x + (x^{2} + 2 x y) d y = 0$ exact? Then solve it.

Recall Solution 3.2

$M = a x y + y^{2}$ , $N = x^{2} + 2 x y$ . Force the test: $M_{y} = a x + 2 y$ , $N_{x} = 2 x + 2 y$ . Match them: $a x + 2 y = 2 x + 2 y$ for all $x, y$ ⟹ $a = 2$ . With $a = 2$ : $M = 2 x y + y^{2}$ . Integrate $M$ : $F = \int (2 x y + y^{2}) d x + g (y) = x^{2} y + x y^{2} + g (y) .$ Match $N$ : $F_{y} = x^{2} + 2 x y + g^{'} (y) = x^{2} + 2 x y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = 0.$ Answer: $a = 2$ and $x^{2} y + x y^{2} = c .$

Exercise 3.3

The equation $(y cos x + 2 x e^{y}) d x + (sin x + x^{2} e^{y} - 1) d y = 0$ claims to be exact. Verify, then find the particular solution through $(0, 0)$ .

Recall Solution 3.3

$M = y cos x + 2 x e^{y}$ , $N = sin x + x^{2} e^{y} - 1$ . Test: $M_{y} = cos x + 2 x e^{y}$ ; $N_{x} = cos x + 2 x e^{y}$ ⟹ exact. ✓ Integrate $M$ : $F = \int (y cos x + 2 x e^{y}) d x + g (y) = y sin x + x^{2} e^{y} + g (y) .$ Match $N$ : $F_{y} = sin x + x^{2} e^{y} + g^{'} (y) = sin x + x^{2} e^{y} - 1 \Rightarrow g^{'} (y) = - 1 \Rightarrow g (y) = - y .$ General solution: $y sin x + x^{2} e^{y} - y = c .$ Apply $(0, 0)$ : $0 \cdot sin 0 + 0 \cdot e^{0} - 0 = c \Rightarrow c = 0.$ Answer: $y sin x + x^{2} e^{y} - y = 0.$

Level 4 — Synthesis

Goal: combine exactness with the "integrate $N$ first" route, and with initial conditions.

Exercise 4.1

Solve $(\frac{1}{y} + 2 x) d x + (- \frac{x}{y ^{2}} + 3 y^{2}) d y = 0$ by integrating $N$ in $y$ first (the symmetric route). State the domain restriction.

Recall Solution 4.1

$M = \frac{1}{y} + 2 x$ , $N = - \frac{x}{y ^{2}} + 3 y^{2}$ . (Requires $y \neq = 0$ .) Test: $M_{y} = - \frac{1}{y ^{2}}$ ; $N_{x} = - \frac{1}{y ^{2}}$ ⟹ exact. ✓ Integrate $N$ in $y$ (treat $x$ as constant), add $h (x)$ this time: $F = \int (- \frac{x}{y ^{2}} + 3 y^{2}) d y + h (x) = \frac{x}{y} + y^{3} + h (x) .$ (Note $\int - x / y^{2} d y = - x \cdot (- 1/ y) = x / y$ .) Match $F_{x} = M$ : $F_{x} = \frac{1}{y} + h^{'} (x) =! \frac{1}{y} + 2 x \Rightarrow h^{'} (x) = 2 x \Rightarrow h (x) = x^{2} .$ Answer: $\frac{x}{y} + y^{3} + x^{2} = c, y \neq = 0.$

Exercise 4.2

Solve the IVP $(2 x y - sec^{2} x) d x + (x^{2} + 2 y) d y = 0$ , $y (0) = 3$ .

Recall Solution 4.2

$M = 2 x y - sec^{2} x$ , $N = x^{2} + 2 y$ . Test: $M_{y} = 2 x$ ; $N_{x} = 2 x$ ⟹ exact. Integrate $M$ : recall $\int sec^{2} x d x = tan x$ , so $F = \int (2 x y - sec^{2} x) d x + g (y) = x^{2} y - tan x + g (y) .$ Match $N$ : $F_{y} = x^{2} + g^{'} (y) = x^{2} + 2 y \Rightarrow g^{'} (y) = 2 y \Rightarrow g (y) = y^{2} .$ General: $x^{2} y - tan x + y^{2} = c .$ Apply $y (0) = 3$ : $0 - tan 0 + 9 = c \Rightarrow c = 9.$ Answer: $x^{2} y - tan x + y^{2} = 9.$

Exercise 4.3

A student integrated $M = 6 x^{2} + 4 x y$ and got $F = 2 x^{3} + 2 x^{2} y + g (y)$ , then matched against $N = 2 x^{2} - 3 y^{2}$ . Complete the solution — and explain, using the picture, why no term of $N$ gets integrated twice.

Recall Solution 4.3

Test first (the student skipped it): $M_{y} = 4 x$ , $N_{x} = 4 x$ ⟹ exact. ✓ Match $N$ : $F_{y} = 2 x^{2} + g^{'} (y) =! 2 x^{2} - 3 y^{2} \Rightarrow g^{'} (y) = - 3 y^{2} \Rightarrow g (y) = - y^{3} .$ Answer: $2 x^{3} + 2 x^{2} y - y^{3} = c .$ Why no double-counting: the part of $N$ that equals $2 x^{2}$ is already produced by differentiating $\int M d x = 2 x^{3} + 2 x^{2} y$ in $y$ . On the hill picture, that slope is already built into $F$ . Only the leftover slope $- 3 y^{2}$ — the part with no $x$ in it — is new information, and it becomes $g^{'} (y)$ . Subtracting it out is exactly step M of I-D-M-I.

Level 5 — Mastery

Goal: reverse-engineer the machinery and handle degenerate / limiting cases.

Exercise 5.1

Design a potential $F (x, y) = x^{3} y^{2}$ . Write the exact ODE it satisfies (in $M d x + N d y = 0$ form), then recover $F$ back from $M$ and $N$ to confirm the round trip.

Recall Solution 5.1

Forward (build the ODE): $d F = F_{x} d x + F_{y} d y$ .

$M = F_{x} = 3 x^{2} y^{2}$ ,
$N = F_{y} = 2 x^{3} y$ .

ODE: $3 x^{2} y^{2} d x + 2 x^{3} y d y = 0.$ Sanity test: $M_{y} = 6 x^{2} y$ , $N_{x} = 6 x^{2} y$ ⟹ exact by construction. ✓ Reverse (recover $F$ ): integrate $M$ in $x$ : $\int 3 x^{2} y^{2} d x + g (y) = x^{3} y^{2} + g (y)$ . Match $N$ : $F_{y} = 2 x^{3} y + g^{'} (y) = 2 x^{3} y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = const$ . Recovered $F = x^{3} y^{2}$ ✓ — the round trip closes. Solution family: $x^{3} y^{2} = c$ .

Exercise 5.2

Consider $(y^{2} - 1) d x + (2 x y) d y = 0$ . Solve it. Then examine the degenerate level $c = 0$ geometrically: what curve(s) does $F (x, y) = 0$ describe?

Recall Solution 5.2

$M = y^{2} - 1$ , $N = 2 x y$ . Test: $M_{y} = 2 y$ , $N_{x} = 2 y$ ⟹ exact. Integrate $M$ : $F = \int (y^{2} - 1) d x + g (y) = x (y^{2} - 1) + g (y) = x y^{2} - x + g (y) .$ Match $N$ : $F_{y} = 2 x y + g^{'} (y) = 2 x y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = 0.$ General solution: $x y^{2} - x = c$ , i.e. $x (y^{2} - 1) = c .$ Degenerate level $c = 0$ : $x (y^{2} - 1) = 0$ splits into $x = 0$ (the $y$ -axis) or $y^{2} = 1$ , i.e. $y = 1$ and $y = - 1$ (two horizontal lines). So the single level curve $F = 0$ is actually a union of three straight lines crossing the plane — a reminder that a contour $F = c$ need not be one smooth curve; it can branch or split where the level passes through a saddle/crossing.

Exercise 5.3

The equation $M d x + N d y = 0$ with $M = 2 x y^{3} + y$ is exact for a suitable $N$ with $N (0, y) = y^{4}$ . Reconstruct $N$ .

Recall Solution 5.3

Exactness demands $N_{x} = M_{y}$ . Compute $M_{y} = \frac{\partial}{\partial y} (2 x y^{3} + y) = 6 x y^{2} + 1.$ So $N_{x} = 6 x y^{2} + 1$ . Integrate in $x$ to recover $N$ (the "constant" is a function of $y$ ): $N = \int (6 x y^{2} + 1) d x + φ (y) = 3 x^{2} y^{2} + x + φ (y) .$ Pin $φ$ using $N (0, y) = y^{4}$ : set $x = 0$ : $N (0, y) = 0 + 0 + φ (y) = y^{4} \Rightarrow φ (y) = y^{4} .$ Answer: $N = 3 x^{2} y^{2} + x + y^{4} .$ (Quick check that a real $F$ exists: $F = x^{2} y^{3} + x y + \frac{1}{5} y^{5}$ gives $F_{x} = 2 x y^{3} + y = M$ and $F_{y} = 3 x^{2} y^{2} + x + y^{4} = N$ . ✓)

Exercise 5.4

Show that every separable equation $p (x) d x + q (y) d y = 0$ is automatically exact. Where does the potential come from?

Recall Solution 5.4

Here $M = p (x)$ depends on $x$ only; $N = q (y)$ depends on $y$ only. Test: $M_{y} = \frac{\partial}{\partial y} p (x) = 0$ (no $y$ inside), and $N_{x} = \frac{\partial}{\partial x} q (y) = 0$ (no $x$ inside). $0 = 0$ ⟹ always exact. ✓ The potential: $F = \int p (x) d x + \int q (y) d y$ , and the solution is $\int p d x + \int q d y = c$ — precisely the answer the separable method gives. So separable equations are a special, "already-split" case of exact ones.

Recall One-line self-check of the whole ladder

If you can (a) run $M_{y} = ? N_{x}$ , (b) integrate either coefficient and patch with $g (y)$ or $h (x)$ , (c) apply an initial condition to fix $c$ , and (d) reverse the process to build an exact equation from a chosen $F$ — you own this topic.