First-order higher-degree ODEs

At a Glance

• Frequency: 5 sub-parts across 4 of 13 years (2015, 2018, 2020, 2025)
• Priority tier: T3
• Marks (count): 10 (3), 13 (2)
• Average solve time: ~13 min
• Difficulty mix: hard 3, medium 2
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom contains the hardest ODE questions on Paper 1 — three of the five appearances are classified hard, and the two 13-mark questions both require a non-standard algebraic manoeuvre before integration begins. The two archetypes split cleanly by the structure of the equation: if you can factor the quadratic in $p$ into a perfect square, or if the equation is linear in $x$ or $y$, you have the right approach. The payoff for learning both techniques is high: the 2025 paper tested both archetypes (Q5a and Q7c-i), for a combined 20 marks.

Minimum Theory

The problem. A first-order ODE of degree $n$ has the form $F(x, y, p) = 0$ where $p = dy/dx$ and $F$ is a polynomial of degree $n > 1$ in $p$. The standard strategies are:

Solvable for $p$: Factor $F$ into linear factors in $p$. Each factor gives a first-order linear ODE; the general solution is the product of individual general solutions.

Solvable for $x$ (Lagrange/d’Alembert type): If the equation can be written as $x = f(p, y)$, differentiate w.r.t. $y$ using $dx/dy = 1/p$ to get a linear ODE in $x$ (or $y$) as a function of $p$.

Solvable for $y$: If $y = f(p, x)$, differentiate w.r.t. $x$ using $dy/dx = p$ to get an ODE in $p$ vs $x$.

Singular solutions. Besides the general solution (containing an arbitrary constant $c$), higher-degree ODEs often have singular solutions — solutions that satisfy the ODE but cannot be obtained from the general solution by any choice of $c$. Geometrically, singular solutions are the envelope of the family of curves described by the general solution.

$p$-discriminant method: From $F(x,y,p)=0$, solve $\partial F/\partial p = 0$ for $p$ and substitute back into $F=0$. The resulting relation in $(x,y)$ gives candidates; test each against the original ODE.

$c$-discriminant method: Eliminate $c$ from the general solution $G(x,y,c)=0$ and $\partial G/\partial c = 0$. Test candidates in the ODE.

Question Archetypes

quadratic-in-p-singular (3 questions; 2020, 2025, 2025)

Recognition Cues

• The ODE is degree 2 in $p$, e.g. $(ap+b)^n = (cp+d)^n$ or $Ap^2 + Bp + C = 0$.
• The question always asks for both the general solution and the singular solution.
• Look for: a perfect-square factorisation, or a missing $x$ (so you can separate directly after solving for $p$), or a useful substitution $v = x + y$.

Solution Template

1. Identify the degree-2 structure. Clear fractions; write as $A(x,y)p^2 + B(x,y)p + C(x,y) = 0$.
2. Try to factor or simplify. Look for perfect squares; group to reveal $(px - y)^2 = (\text{something positive})$.
3. Separate and integrate. Once $p$ is isolated (or as a substitution), separate variables and integrate. Use standard substitutions ($u = 3-y$, $u = x/y$, $v = x+y$) to clean up the integrand.
4. State the general solution. Implicit or explicit.
5. Find the singular solution. Use the $p$-discriminant: set $\partial F/\partial p = 0$, solve for $p$, substitute into $F = 0$. Test each candidate against the original ODE — not all solutions of the discriminant equation are genuine singular solutions.

Worked Example(s)

2020 Paper 1, 2020-P1-Q8a-ii (10 marks)

Find general and singular solutions of $9p^2(2-y)^2 = 4(3-y)$.

Solve for $p$: $p = \pm\dfrac{2}{3}\dfrac{\sqrt{3-y}}{2-y}$. Separate (taking the $+$ branch): $dx = \frac{3}{2}\,\frac{(2-y)}{\sqrt{3-y}}\,dy.$ Substitute $u = 3-y$: $dx = -\tfrac{3}{2}(u-1)/\sqrt{u}\,du$. Integrate: $x = -u^{3/2} + 3\sqrt{u} + C \;\Rightarrow\; x - C = y\sqrt{3-y}.$

General solution: $(x-C)^2 = y^2(3-y).$

Singular solution. $p$-discriminant: $F = 9(2-y)^2p^2 - 4(3-y)$; $\partial F/\partial p = 18(2-y)^2 p = 0 \Rightarrow p = 0$ or $y = 2$.

• $p = 0 \Rightarrow 4(3-y) = 0 \Rightarrow y = 3$. Check: $9(0)(2-3)^2 = 0 = 4(0)$ ✓.
• $y = 2$: $9p^2(0) = 4(1) \Rightarrow 0 = 4$. Rejected.

$\boxed{\text{General: } (x-C)^2 = y^2(3-y);\quad \text{Singular: } y = 3.}$

2025 Paper 1, 2025-P1-Q5a (10 marks)

Solve $\left(1 - y^2 + \dfrac{y^4}{x^2}\right)p^2 - \dfrac{2y}{x}p + \dfrac{y^2}{x^2} = 0$.

Multiply by $x^2$: $(x^2 - x^2y^2 + y^4)p^2 - 2xyp + y^2 = 0$.

Key factorisation: $(px-y)^2 = p^2y^2(x^2-y^2)$, i.e. $px - y = \pm py\sqrt{x^2-y^2}$.

Rearrange: $p(x \mp y\sqrt{x^2-y^2}) = y$. Divide both sides and use $u = x/y$, $du = (x\,dy - y\,dx)/y^2$: $\frac{du}{\sqrt{u^2-1}} = \mp dy \;\Rightarrow\; \cosh^{-1}(x/y) = \mp y + \text{const}.$

General solution: $x = y\cosh(y + c)$.

Singular solution. The discriminant $B^2 - 4AC = 4y^4(x^2-y^2)/x^4 = 0$ when $x = \pm y$. Both satisfy the original ODE (direct check with $p = \pm 1$).

$\boxed{\text{General: } x = y\cosh(y+c);\quad \text{Singular: } y = \pm x.}$

2025 Paper 1, 2025-P1-Q7c-i (10 marks)

General and singular solutions of $(1+p)^3 = \dfrac{27}{8a}(x+y)(1-p)^3$.

Substitution: $v = x + y$, so $dv/dx = 1 + p$. After taking cube roots and solving for $v' = dv/dx$: $dx = \left(\frac{a^{1/3}}{3}v^{-1/3} + \frac{1}{2}\right)dv.$ Integrate and substitute back $v = x+y$: cube both sides to clear the fractional power.

General solution: $(x - y + c)^3 = a(x+y)^2$.

Singular solution. $\partial F/\partial p = 0$ gives $p = -1$, then $x + y = 0$.

$\boxed{\text{General: } (x-y+c)^3 = a(x+y)^2;\quad \text{Singular: } x+y=0.}$

Common Traps

• Always test $p$-discriminant candidates. Not every root of $\partial F/\partial p = 0$ is a singular solution: the value must also satisfy the original ODE. In the 2020 example, $y=2$ fails the ODE test.
• The $c$-discriminant can produce spurious factors. In the 2020 example, differentiating $(x-C)^2 = y^2(3-y)$ w.r.t. $C$ gives $C = x$, substituting yields $y^2(3-y) = 0$: the factors $y = 0$ (a node locus) and $y = 3$ (the envelope). Only $y = 3$ satisfies the ODE.
• Singular solutions are NOT obtained from the general solution for any value of $c$. Verify this is the case before declaring a solution “singular”.
• The substitution $u = 3-y$ (2020) or $u = x/y$ (2025-Q5a) or $v = x+y$ (2025-Q7c-i) is the key step that linearises the ODE for integration; practice recognising the right substitution.

solvable-for-x (2 questions; 2015, 2018)

Recognition Cues

• The equation can be written as $x = f(p, y)$ (solvable for $x$, or equivalently linear in $x$).
• The equation is degree 2 in $p$ but linear in $x$, making direct solution for $p$ messy but solution for $x$ clean.
• Key phrase: “solvable for $x$” or “Lagrange/d’Alembert form”.

Solution Template

1. Solve for $x$ (if the equation is linear in $x$): $x = g(y, p)$.
2. Differentiate w.r.t. $y$, using $dx/dy = 1/p$: $\frac{1}{p} = \frac{\partial g}{\partial y} + \frac{\partial g}{\partial p}\frac{dp}{dy}.$
3. Separate or recognise the linear ODE in $(y, p)$ or $(x, p)$. Solve for $yp = c$ (or similar) by separation.
4. Eliminate $p$ from the relation $yp = c$ (or equivalent) and the original $x = g(y,p)$.

Worked Example(s)

2018 Paper 1, 2018-P1-Q6a (13 marks)

Solve $p^2 y + 2px - y = 0$.

The equation is linear in $x$: $2px = y - p^2y = y(1-p^2)$, so $x = \dfrac{y(1-p^2)}{2p}$.

Differentiate w.r.t. $y$, using $dx/dy = 1/p$: $\frac{1}{p} = \frac{1-p^2}{2p} - \frac{y(1+p^2)}{2p^2}\frac{dp}{dy} \cdot p.$ After algebra (move $\tfrac{1}{p} - \tfrac{1-p^2}{2p} = \tfrac{1+p^2}{2p}$ to the left): $\frac{dy}{y} = -\frac{dp}{p} \;\Rightarrow\; yp = c.$

Substitute $p = c/y$ into $x = y(1-p^2)/(2p)$: $x = \frac{y(1 - c^2/y^2)}{2c/y} = \frac{y^2 - c^2}{2c}.$ $\boxed{y^2 = 2cx + c^2 \qquad (c \text{ arbitrary}).}$ This is a family of parabolas with axis along the $x$-axis.

2015 Paper 1, 2015-P1-Q7d (13 marks)

Solve $x = py - p^2$.

Differentiate w.r.t. $y$: $(1-p^2)\,dy/dp = py - 2p^2$, giving: $\frac{dy}{dp} - \frac{p}{1-p^2}\,y = \frac{-2p^2}{1-p^2}.$ Integrating factor $= \sqrt{1-p^2}$. After multiplying and integrating (using $p = \sin\theta$): $y\sqrt{1-p^2} = p\sqrt{1-p^2} - \sin^{-1}p + C.$

Parametric general solution: $y = p + \frac{C - \sin^{-1}p}{\sqrt{1-p^2}},\qquad x = py - p^2,\quad p \in (-1, 1).$

Singular solutions from $1 - p^2 = 0$, i.e. $p = \pm 1$: the lines $y = \pm(x+1)$.

Common Traps

• Recognise linearity in $x$, not $p$. Trying to solve the 2018 ODE as a quadratic in $p$ gives $p = (-2x \pm \sqrt{4x^2+4y^2})/(2y)$ — messy. The $x$-linearity route gives clean separation.
• The factor $\tfrac{1+p^2}{2}$ cancels in the 2018 derivation; write it explicitly before cancelling.
• The 2015 integral $\int p^2/\sqrt{1-p^2}\,dp$ requires $p = \sin\theta$; the result is $\tfrac{1}{2}(\sin^{-1}p - p\sqrt{1-p^2})$.
• Final answer is parametric in $p$ for the 2015 equation — eliminating $p$ in closed form is not expected.

Marks-Aware Writing

10-mark singular-solution question: State the general solution (derived clearly) in 4–5 lines; then apply the discriminant method in 3 lines. Explicitly test each discriminant candidate against the original ODE. Marks are evenly split between general and singular.

13-mark solvable-for-$x$ question: The Lagrange differentiation step must be written out in full (2–3 lines). Show that the ODE reduces to a separable equation; state the separation explicitly. Derive the parametric or explicit general solution; state singular solutions if they arise. The examiner awards marks at each step.

Practice Set

• 2023-P1-Q7a-ii (10 m) —
• 2022-P1-Q8a-i (10 m) —
• 2021-P1-Q7b (15 m) —
• 2019-P1-Q8a (15 m) —
• 2017-P1-Q6b-i (10 m) —