Exact equations

At a Glance

• Frequency: 9 sub-parts across 6 of 13 years (2013, 2014, 2015, 2018, 2019, 2023)
• Priority tier: T2
• Marks (count): 10 (5), 12 (3), 15 (1)
• Average solve time: ~9 min
• Difficulty mix: medium 6, easy 3
• Section: B | Dominant type: computation

Why This Chapter Matters

Exact and integrating-factor questions appear in 6 of the last 13 years and are mostly 10–12 marks in Section B — short enough to be attempted under time pressure. Eight of the nine questions require finding an integrating factor; one is directly exact. The standard type (IF depending only on $x$ or only on $y$) is three or four minutes of formula application once you recognise the form. The harder variants — IF of the form $x^\alpha y^\beta$, or a function of $x+y$, or finding $f(y)$ for exactness — are less mechanical but still fully algorithmic. Every question reduces to the same three steps: test exactness, identify the IF, integrate the potential.

Minimum Theory

Exactness test. $M(x,y)\,dx+N(x,y)\,dy=0$ is exact iff $\partial M/\partial y = \partial N/\partial x$ throughout a simply-connected domain.

Integrating factor (IF). Multiply through by $\mu$ to make the equation exact. The IF is found by the form it takes:

• $\mu=\mu(x)$ only: requires $(M_y-N_x)/N$ to be a function of $x$ alone; then $\mu(x)=\exp\!\int\!\tfrac{M_y-N_x}{N}\,dx$.
• $\mu=\mu(y)$ only: requires $(N_x-M_y)/M$ to be a function of $y$ alone; then $\mu(y)=\exp\!\int\!\tfrac{N_x-M_y}{M}\,dy$.
• $\mu=\mu(x+y)$: requires $(N_x-M_y)/(M-N)$ to be a function of $x+y$ alone; then $\mu=\exp\!\int f(x+y)\,d(x+y)$.
• $\mu=x^\alpha y^\beta$: multiply through, impose exactness, match coefficients of monomials to get a $2\times2$ linear system for $\alpha,\beta$.

Integrating the potential. Once exact, find $F(x,y)$ with $F_x=M,\,F_y=N$ by integrating one partial, then matching the other to find $g(\cdot)$. General solution: $F(x,y)=C$.

Exactness-condition type. Given an equation with an unknown function, impose $M_y=N_x$ to find it, then solve the now-exact equation.

Question Archetypes

Three patterns cover every exact/IF question in the corpus.

integrating-factor (7 question(s); 2013, 2014, 2014, 2015, 2015, 2018, 2019)

Recognition Cues

• “Solve $M\,dx+N\,dy=0$” where $M_y\ne N_x$ — you must find an IF.
• “Show the equation is non-exact; find an IF of the form […].”
• Variant: “Find $\alpha,\beta$ such that $x^\alpha y^\beta$ is an IF.”
• Variant: “Find the sufficient condition for an IF as a function of $(x+y)$; apply to […].”

Solution Template

1. Test exactness. Compute $M_y$ and $N_x$; confirm they differ.
2. Identify the IF form. In order: try $(M_y-N_x)/N$ (pure $x$?); try $(N_x-M_y)/M$ (pure $y$?); try $(N_x-M_y)/(M-N)$ (pure $x+y$?); else try $x^\alpha y^\beta$.
3. Compute the IF. Apply the appropriate integral/formula.
4. Multiply and verify. After multiplying by $\mu$, recheck $M^\ast_y=N^\ast_x$.
5. Integrate. Find $F(x,y)$: integrate $M^\ast$ in $x$ to get $F=\ldots+g(y)$; differentiate in $y$, match $N^\ast$ to find $g(y)$.
6. Write $F=C$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q6a (10 marks)

Solve $(5x^3+12x^2+6y^2)\,dx+6xy\,dy=0$.

$M_y=12y$, $N_x=6y$. Not exact. $(M_y-N_x)/N=(12y-6y)/(6xy)=1/x$ — pure $x$. So $\mu=e^{\int(1/x)dx}=x$.

After multiplying by $x$: $(5x^4+12x^3+6xy^2)\,dx+6x^2y\,dy=0$. Now $M'_y=12xy=N'_x$ ✓.

Integrate: $F=\int6x^2y\,dy=3x^2y^2+g(x)$. Match $F_x=5x^4+12x^3+6xy^2$: $g'(x)=5x^4+12x^3$, $g=x^5+3x^4$.

$\boxed{x^5+3x^4+3x^2y^2=C.}$

2014 Paper 1, 2014-P1-Q8a (15 marks)

State sufficient condition for IF as function of $(x+y)$; apply to $(x^2+xy)\,dx+(y^2+xy)\,dy=0$.

Condition: $(N_x-M_y)/(M-N)$ must be a function of $(x+y)$ only.

Here $M_y=x$, $N_x=y$. $(N_x-M_y)/(M-N)=(y-x)/(x^2+xy-y^2-xy)=(y-x)/(x^2-y^2)=-1/(x+y)$ — pure $(x+y)$ ✓.

$\mu=\exp(-\ln|x+y|)=1/(x+y)$. Multiplying: $x\,dx+y\,dy=0$ (the $(x+y)$ cancels). Integrate:

$\boxed{x^2+y^2=C.}$

2018 Paper 1, 2018-P1-Q7d (12 marks)

Find $\alpha,\beta$ such that $x^\alpha y^\beta$ is an IF of $(4y^2+3xy)\,dx-(3xy+2x^2)\,dy=0$.

Multiply $M\,dx+N\,dy=0$ by $x^\alpha y^\beta$. Impose exactness; match coefficients of $x^\alpha y^{\beta+1}$ and $x^{\alpha+1}y^\beta$: $3\alpha+4\beta=-11,\quad 2\alpha+3\beta=-7.$ Solve: $\alpha=-5$, $\beta=1$. With $\mu=x^{-5}y$: integrate the exact equation to get

$\boxed{\frac{y^2}{x^3}+\frac{y^3}{x^4}=C.}$

2019 Paper 1, 2019-P1-Q5a (10 marks)

Solve $(2y\sin x+3y^4\sin x\cos x)\,dx-(4y^3\cos^2 x+\cos x)\,dy=0$.

$M_y-N_x=\sin x(1+8y^3\cos x)-(4y^3\sin x\cos x+\sin x)$ … testing $(N_x-M_y)/M$ simplifies to $-\tan x$ — pure $x$. Wait: $(M_y-N_x)/N=(-\tan x)$ for this case (verify). So $\mu=\cos x$.

After multiplying by $\cos x$: integrate $N^\ast=-4y^3\cos^3 x-\cos^2 x$ in $y$: $F=-y^4\cos^3 x-y\cos^2 x+h(x)$. Match $F_x=M^\ast$: $h'(x)=0$.

$\boxed{y^4\cos^3 x+y\cos^2 x=C.}$

Common Traps

• Test $(M_y-N_x)/N$ for $x$-IF and $(N_x-M_y)/M$ for $y$-IF — the signs are opposite. Swapping them gives the wrong formula and a wrong IF.
• $(x+y)$ condition uses $M-N$ in the denominator, not $N$ or $M$ alone. The numerator is $N_x-M_y$ (note: same sign as the $y$-only formula’s numerator).
• For $x^\alpha y^\beta$: expand $M^\ast$ and $N^\ast$ fully before matching coefficients. Keep track of whether $N$ has a minus sign in the original equation.
• After multiplying by $\mu$, always recheck $M^\ast_y=N^\ast_x$ before integrating — a sign error in the IF is easy to miss otherwise.

exact-equation (1 question(s); 2023)

Recognition Cues

• Test exactness and it passes: $M_y=N_x$ immediately.
• Often signals itself by having clean cross-partials that are obviously equal.

Solution Template

1. Verify: compute $M_y$ and $N_x$, confirm equality.
2. Integrate $F_x=M$ in $x$: $F=\int M\,dx+g(y)$.
3. Match $F_y=N$: differentiate $F$ in $y$, equate to $N$, solve for $g'(y)$, integrate.
4. Write $F=C$.

Worked Example(s)

2023 Paper 1, 2023-P1-Q7a-i (10 marks)

Solve $dy/dx=-(2xy^3+2)/(3x^2y^2+8e^{4y})$.

Rewrite: $(2xy^3+2)\,dx+(3x^2y^2+8e^{4y})\,dy=0$. $M_y=6xy^2=N_x$ — directly exact.

$F=\int(2xy^3+2)\,dx=x^2y^3+2x+g(y)$. $F_y=3x^2y^2+g'(y)=3x^2y^2+8e^{4y}$, so $g'(y)=8e^{4y}$, $g=2e^{4y}$.

$\boxed{x^2y^3+2x+2e^{4y}=C.}$

Common Traps

• Always test exactness first — do not assume non-exactness just because the equation looks complicated.
• The $e^{4y}$ term in $N$ integrates to $2e^{4y}$, not $e^{4y}/4$. Track the constant carefully.

exactness-condition (1 question(s); 2018)

Recognition Cues

• “Find $f(y)$ [or $f(x)$, or parameter $a$] such that $M\,dx+N\,dy=0$ is exact.”
• One of $M,N$ contains an unknown function; solve for it using $M_y=N_x$, then integrate.

Solution Template

1. Apply $M_y=N_x$: differentiate both sides partially; the unknown function’s derivative appears directly.
2. Integrate to find the unknown function (take the simplest antiderivative).
3. Solve the now-exact equation using the standard template.

Worked Example(s)

2018 Paper 1, 2018-P1-Q8d (12 marks)

Find $f(y)$ so $(2xe^y+3y^2)\,dy+(3x^2+f(y))\,dx=0$ is exact; then solve.

$P=3x^2+f(y)$, $Q=2xe^y+3y^2$. Exactness: $P_y=f'(y)$, $Q_x=2e^y$. So $f'(y)=2e^y$, giving $f(y)=2e^y$.

Now integrate: $F=\int P\,dx=x^3+2xe^y+g(y)$. $F_y=2xe^y+g'(y)=Q=2xe^y+3y^2$, so $g=y^3$.

$\boxed{x^3+2xe^y+y^3=C.}$

Common Traps

• Identify $P$ (coefficient of $dx$) and $Q$ (coefficient of $dy$) correctly — the equation may be written with the $dy$ term first.
• An additive constant in $f(y)$ is harmless and absorbed into $C$.

Practice Set

• 2014-P1-Q5a (10 m) — — show $1/(x^2+y^2)$ is an IF for a given equation; verify and solve.
• 2015-P1-Q6a (12 m) — — find the value of $a$ such that $(x+y)^a$ is an IF; solve.