Quasilinear first-order PDEs (Lagrange’s method)

At a Glance

• Frequency: 9 sub-parts across 7 of 13 years (2013, 2015, 2016, 2018, 2019, 2020, 2024)
• Priority tier: T2
• Marks (count): 10 (3), 15 (6)
• Average solve time: ~14 min
• Difficulty mix: hard 5, medium 4
• Section: B | Dominant type: computation

Why This Chapter Matters

Quasilinear first-order PDEs appear in 7 of the last 12 years, split evenly between 10- and 15-mark slots and rated the hardest PDE atom in Paper 2. The difficulty is almost entirely in finding the right multipliers for Lagrange’s auxiliary equations — once the two first integrals are found, the general solution is immediate. Three question types repeat: pure Lagrange (find two first integrals, write the general solution), integral surface through a curve (additionally eliminate a parameter to pin down a specific surface), and orthogonal trajectories (form the PDE from the family’s gradient and apply Lagrange). The multiplier patterns that recur — $(x,y,z)$, $(y,x,0)$, $(3x^2,3y^2,0)$, and “add all three ratios” — are worth memorising.

Minimum Theory

Lagrange’s quasilinear PDE. The PDE $P(x,y,z)p+Q(x,y,z)q=R(x,y,z)$ (where $p=z_x$, $q=z_y$) has general solution $F(u_1,u_2)=0$ where $u_1=c_1$ and $u_2=c_2$ are two independent first integrals of the Lagrange auxiliary (subsidiary) equations: $\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}.$

Finding first integrals via multipliers. For a common ratio $k=dx/P=dy/Q=dz/R$, any multipliers $(\ell,m,n)$ give $k=(\ell\,dx+m\,dy+n\,dz)/(\ell P+mQ+nR)$. If $\ell P+mQ+nR=0$ then $\ell\,dx+m\,dy+n\,dz=0$ along characteristics, giving an exact first integral when the numerator is integrable. Good multiplier choices: $(x,y,z)$ (looks for $x^2+y^2+z^2$ structure); $(y,x,0)$ (gives $d(xy)$); $(1,1,1)$ (gives $x+y+z$); $(x,y,\phi)$ (gives $x^2+y^2+\phi^2$).

Integral surface through a curve. Given parametric curve $x=x_0(t),\;y=y_0(t),\;z=z_0(t)$: evaluate $c_1=u_1(x_0,y_0,z_0)$ and $c_2=u_2(x_0,y_0,z_0)$ as functions of the parameter $t$; eliminate $t$ to get the relation $F(c_1,c_2)=0$ that the integral surface satisfies.

Orthogonal trajectory surfaces. For a one-parameter family $\phi(x,y,z)=C$, the normal direction at each point is $\nabla\phi$; orthogonal surfaces satisfy the PDE $\phi_x p+\phi_y q=\phi_z$ (after eliminating $C$). Solve by Lagrange’s method.

Question Archetypes

Three patterns cover every quasilinear PDE question in the corpus.

integral-surface (4 question(s); 2018, 2019, 2020, 2024)

Recognition Cues

• A Lagrange PDE is given and a curve is specified (“passes through $x=t,y=t^2,z=1$” or “integral surface containing $xz=a^3,y=0$”).
• The phrase “integral surface through” or “that contains the curve.”
• Answering requires two steps: find the general solution, then apply the curve to pin down the specific surface.

Solution Template

1. Write the auxiliary equations $dx/P=dy/Q=dz/R$.
2. Find two independent first integrals $u_1=c_1$ and $u_2=c_2$ using multipliers or direct pairing.
3. Write the general solution $F(u_1,u_2)=0$ (or $u_2=G(u_1)$ equivalently).
4. On the given curve, compute $c_1(t)=u_1(x_0(t),y_0(t),z_0(t))$ and $c_2(t)=u_2(x_0(t),y_0(t),z_0(t))$.
5. Eliminate the parameter $t$ between $c_1(t)$ and $c_2(t)$ to find the relation $F(c_1,c_2)=0$.
6. Substitute back $u_1,u_2$ to write the surface equation explicitly.

Worked Example(s)

2018 Paper 2, 2018-P2-Q6a (15 marks)

$(y^3x-2x^4)p+(2y^4-x^3y)q=9z(x^3-y^3)$; integral surface through $x=t,y=t^2,z=1$.

Auxiliary equations: $\dfrac{dx}{x(y^3-2x^3)}=\dfrac{dy}{y(2y^3-x^3)}=\dfrac{dz}{9z(x^3-y^3)}$.

First integral (multipliers $(y,x,0)$): $yP+xQ=-3xy(x^3-y^3)$; numerator $=d(xy)$. Pairing with $dz/R$ and cancelling $(x^3-y^3)$: $d(xy)/(-3xy)=dz/(9z)$, integrating gives $u_1=z(xy)^3=c_1$.

Second integral (multipliers $(3x^2,3y^2,0)$): denominator $=-6(x^3-y^3)(x^3+y^3)$; numerator $=d(x^3+y^3)$. Same cancel gives $u_2=z^2(x^3+y^3)^3=c_2$.

General solution: $\Phi(z(xy)^3,\;z^2(x^3+y^3)^3)=0$.

Integral surface. On the curve $(x,y,z)=(t,t^2,1)$: $c_1=t^9$, $c_2=(t^3+t^6)^3=t^9(1+t^3)^3$. Eliminate $t$: $c_1^{1/3}=t^3$, then $c_2^{1/3}=t^3(1+t^3)=c_1^{1/3}(1+c_1^{1/3})$. After cubing:

$\boxed{\;z(x^3+y^3-x^2y^2)^3=(xy)^3.\;}$

2019 Paper 2, 2019-P2-Q6a (15 marks)

$xu_x+(u-x-y)u_y=x+2y$; $u=1+y$ on $x=1$.

Auxiliary: $\dfrac{dx}{x}=\dfrac{dy}{u-x-y}=\dfrac{du}{x+2y}$.

First integral: add $dy+du$: numerator $d(u+y)$, denominator $(u+y)$. Pairing with $dx/x$: $d(u+y)/(u+y)=dx/x$, giving $C_1=(u+y)/x$.

Second integral: along characteristic with $x=e^s$, $y$ satisfies the linear ODE $y'+2y=(C_1-1)e^s$; solution $y=A/x^2+(C_1-1)x/3$. Then $C_2=x^2(x+2y-u)$.

Applying $u=1+y$ on $x=1$: $C_1=1+2y$, $C_2=y$. Eliminate $y$: $C_1=1+2C_2$. Solving for $u$:

$\boxed{\;u(x,y)=\dfrac{2x^4+4x^3y+x-y}{1+2x^3}.\;}$

2020 Paper 2, 2020-P2-Q6a (15 marks)

$(x-y)y^2z_x+(y-x)x^2z_y=(x^2+y^2)z$; integral surface through $xz=a^3,y=0$.

First integral: cancel the shared factor $(x-y)$ from $P,Q$: $dx/y^2=dy/(-x^2)$, giving $x^2\,dx+y^2\,dy=0$, so $u_1=x^3+y^3=c_1$.

Second integral: test $v=z/(x-y)$; verify $P v_x+Q v_y+R v_z=0$ ✓, so $u_2=z/(x-y)=c_2$.

Applying curve $y=0$, $z=a^3/x$: $c_1=x^3$, $c_2=z/x=a^3/x^2$. Eliminate $x$: $c_1=x^3\Rightarrow x=c_1^{1/3}$, so $c_2 c_1^{2/3}=a^3$, i.e. $c_2^3c_1^2=a^9$.

$\boxed{\;(x^3+y^3)^2 z^3=a^9(x-y)^3.\;}$

2024 Paper 2, 2024-P2-Q7a (15 marks)

$(y-\phi)\phi_x+(\phi-x)\phi_y=x-y$; through $\phi=0,xy=1$ and circle $x+y+\phi=0,x^2+y^2+\phi^2=a^2$.

First integral: add all three ratios — denominator $=(y-\phi)+(\phi-x)+(x-y)=0$, so $dx+dy+d\phi=0$, giving $C_1=x+y+\phi$.

Second integral: multiply numerators by $x,y,\phi$: denominator $x(y-\phi)+y(\phi-x)+\phi(x-y)=0$, so $x\,dx+y\,dy+\phi\,d\phi=0$, giving $C_2=x^2+y^2+\phi^2$.

General solution: $x^2+y^2+\phi^2=G(x+y+\phi)$.

Through $\phi=0,xy=1$: $C_1=x+y$, $C_2=x^2+y^2=(x+y)^2-2xy=C_1^2-2$. So $G(t)=t^2-2$.

Surface: $x^2+y^2+\phi^2=(x+y+\phi)^2-2$. Expand and simplify:

$\boxed{\;xy+y\phi+x\phi=1.\;}$

(The second prescribed condition — the circle — gives $a^2=-2$, impossible for real $a$; flag this inconsistency.)

Common Traps

• The shared factor in the denominator of two ratios (e.g. $(x^3-y^3)$ in 2018) cancels against $R$, making the resulting ODE integrable. Look for this cancellation before trying brute-force pairings.
• Eliminate $t$ from the data curve, not from a generic point on the surface. The relation between $c_1(t)$ and $c_2(t)$ is specific to the curve; don’t try to express it in terms of $x,y,z$ directly.
• In 2024, both “add all three” and “multiply by $x,y,\phi$” both give zero denominator — this is a special structure (characteristic condition) that yields two clean integrals immediately. When the denominator vanishes, the numerator equals zero along characteristics.

lagrange-linear (3 question(s); 2015, 2015, 2016)

Recognition Cues

• A Lagrange PDE is given with no prescribed curve; just “find the general integral/solution.”
• The phrase “general integral” or “general solution” (not “integral surface”).
• Often a 10-mark question from Section B compulsory part.

Solution Template

1. Write auxiliary equations $dx/P=dy/Q=dz/R$.
2. Find the two most accessible first integrals by: (a) pairing two ratios directly ($dy/y=dz/z$ etc.); (b) using multipliers to build exact differentials.
3. State the general integral $F(u_1,u_2)=0$ (or $u_2=f(u_1)$).

Worked Example(s)

2015 Paper 2, 2015-P2-Q5a (10 marks)

$(y^2+z^2-x^2)p-2xyq+2xz=0$.

Standard form: $Pp+Qq=R$ with $R=-2xz$. Auxiliary: $\dfrac{dx}{y^2+z^2-x^2}=\dfrac{dy}{-2xy}=\dfrac{dz}{-2xz}$.

First integral: pair $dy/(-2xy)=dz/(-2xz)$: $dy/y=dz/z$, giving $u_1=y/z=c_1$.

Second integral: multipliers $(x,y,z)$: $xP+yQ+zR=xz^2+xy^2-x^3-2xy^2-2xz^2=-x(x^2+y^2+z^2)$; numerator $x\,dx+y\,dy+z\,dz=\tfrac{1}{2}d(x^2+y^2+z^2)$. Pairing with $dz/(-2xz)$ and cancelling $-x$: gives $u_2=(x^2+y^2+z^2)/z=c_2$.

$\boxed{\;\Phi\!\left(\frac{y}{z},\;\frac{x^2+y^2+z^2}{z}\right)=0.\;}$

2015 Paper 2, 2015-P2-Q6a (15 marks)

$p\cos(x+y)+q\sin(x+y)=z$.

Let $u=x+y$, $v=x-y$. Auxiliary: $du/(\cos u+\sin u)=dv/(\cos u-\sin u)=dz/z$.

First integral (from $du$ and $dz$): integrate $dz/z=du/(\cos u+\sin u)$. Use $\cos u+\sin u=\sqrt{2}\sin(u+\pi/4)$; $\int\csc v\,dv=\ln|\tan(v/2)|$: $C_1=\frac{[\tan((x+y)/2+\pi/8)]^{1/\sqrt{2}}}{z}.$

Second integral (from $du$ and $dv$): $dv/du=(\cos u-\sin u)/(\cos u+\sin u)=d(\ln|\cos u+\sin u|)/du$, so: $C_2=x-y-\ln|\sin(x+y)+\cos(x+y)|.$

$\boxed{\;\Phi\!\Big(C_1,\;x-y-\ln|\sin(x+y)+\cos(x+y)|\Big)=0.\;}$

2016 Paper 2, 2016-P2-Q5e (10 marks)

$(y+zx)p-(x+yz)q=x^2-y^2$.

First integral (multipliers $(x,y,z)$): $xP+yQ+zR=xz^2-yz^2+zx^2-zy^2+z(x^2-y^2)=$ simplifies to $z(x^2-y^2)=zR$. Numerator $x\,dx+y\,dy+z\,dz=\tfrac12 d(x^2+y^2+z^2)$; relation $d(x^2+y^2-z^2)=0$ gives $u_1=x^2+y^2-z^2=c_1$.

Wait — re-examine: $xP+yQ+zR=x(y+zx)-y(x+yz)+z(x^2-y^2)=x^2z-y^2z=z(x^2-y^2)=zR$. So $\frac{x\,dx+y\,dy+z\,dz}{z(x^2-y^2)}=\frac{dz}{x^2-y^2}$, giving $x\,dx+y\,dy=z\,dz-z\,dz=0$. Actually more directly: $x\,dx+y\,dy+z\,dz=z\,dz$, i.e. $x\,dx+y\,dy-z\,dz=0$, integrating to $u_1=x^2+y^2-z^2$.

Second integral (multipliers $(y,x,1)$): $yP+xQ+R=(y^2-x^2)=-R$. So $y\,dx+x\,dy=-dz$, giving $d(xy+z)=0$, $u_2=xy+z=c_2$.

$\boxed{\;F(x^2+y^2-z^2,\;xy+z)=0.\;}$

Common Traps

• For the 2016 equation, both pairs of multipliers $(x,y,z)$ and $(y,x,1)$ produce something proportional to $R=x^2-y^2$ in the denominator, which is the signature of a clean integral. Test these standard multiplier sets first.
• In $p\cos(x+y)+q\sin(x+y)=z$ (2015), the change of variables $u=x+y$, $v=x-y$ decouples the second integral: $dv/du=(\cos u-\sin u)/(\cos u+\sin u)$ is the derivative of $\ln|\cos u+\sin u|$.

orthogonal-surfaces (2 question(s); 2013, 2016)

Recognition Cues

• A one-parameter family of surfaces $F(x,y,z)=C$ is given; find the orthogonal trajectory surfaces.
• The phrase “surfaces orthogonal to the family” or “orthogonal trajectory surface of the system.”

Solution Template

1. Eliminate the parameter $C$ from $F=C$ by differentiating partially or substituting $C$ as a function of $x,y,z$.
2. Compute the normal direction $(\phi_x,\phi_y,\phi_z)$ of the family (after eliminating $C$).
3. The orthogonal surface satisfies $\phi_x p+\phi_y q=\phi_z$ (or $\phi_x\,dx=\phi_y\,dy=\phi_z\,dz$, equivalently the surface contains the family’s normal).
4. Write Lagrange’s auxiliary: $dx/\phi_x=dy/\phi_y=dz/\phi_z$.
5. Find two first integrals and write the general surface $F(u_1,u_2)=0$.
6. If a specific surface is required, apply an initial curve condition.

Worked Example(s)

2013 Paper 2, 2013-P2-Q6b (15 marks)

Orthogonal to $z(x+y)=C(3z+1)$; through circle $x^2+y^2=1,z=1$.

Family: $\phi=z(x+y)/(3z+1)=C$. Normal: $\phi_x=\phi_y=z/(3z+1)$, $\phi_z=(x+y)/(3z+1)^2$.

Auxiliary (multiply through by $(3z+1)$): $\dfrac{dx}{z}=\dfrac{dy}{z}=\dfrac{dz(3z+1)}{x+y}$.

First integral: $dx=dy$ → $u_1=x-y=c_1$.

Second integral: with $y=x-c_1$, $x+y=2x-c_1$; cross-multiply: $(2x-c_1)\,dx=(3z^2+z)\,dz$. Integrate: $x^2-c_1x=z^3+z^2/2+c_2$, i.e. $u_2=xy-z^3-z^2/2=c_2$.

Through $x^2+y^2=1,z=1$: $c_1=x-y$, $c_2=xy-3/2$. Using $(x-y)^2=1-2xy$: $xy=(1-c_1^2)/2$, so $c_2=(1-c_1^2)/2-3/2=-1-c_1^2/2$, i.e. $c_1^2+2c_2+2=0$.

Substituting back: $\boxed{\;x^2+y^2=2z^3+z^2-2.\;}$

2016 Paper 2, 2016-P2-Q5a (10 marks)

Surfaces orthogonal to $x^2+y^2+z^2=cz$.

Eliminate $c=(x^2+y^2+z^2)/z$: normal $=(2x,2y,(z^2-x^2-y^2)/z)$, equivalently $(2xz,2yz,z^2-x^2-y^2)$.

Auxiliary: $\dfrac{dx}{2xz}=\dfrac{dy}{2yz}=\dfrac{dz}{z^2-x^2-y^2}$.

First integral: $dx/x=dy/y$ → $u_1=y/x=c_1$.

Second integral (multipliers $(x,y,z)$): numerator $x\,dx+y\,dy+z\,dz=\tfrac12 d(r^2)$ where $r^2=x^2+y^2+z^2$; denominator $z(x^2+y^2+z^2)=zr^2$. Pairing with $dy/(2yz)$: $\frac{d(r^2)}{2r^2}=\frac{dy}{2y}$, integrating to $u_2=(x^2+y^2+z^2)/y=c_2$.

$\boxed{\;\Phi\!\left(\frac{y}{x},\;\frac{x^2+y^2+z^2}{y}\right)=0.\;}$

Common Traps

• Eliminate $C$ before forming the normal. The family $x^2+y^2+z^2=cz$ has $c=(r^2)/z$; only after substituting does the third component of the normal become $z^2-x^2-y^2$.
• For 2013, the two integrals are $x-y$ and $xy-z^3-z^2/2$; pinning the second integral down requires solving an ODE in $x$ (or $y$) after substituting $c_1=x-y$ — don’t forget this step.

Marks-Aware Writing

10-mark questions (2015-Q5a, 2016-Q5a, 2016-Q5e): Two to three lines of multiplier setup, one boxed general solution. Show the multiplier calculation (denominator $=0$, numerator $=d(\ldots)$) explicitly — that is the method step worth marks.

15-mark questions (2013, 2015-Q6a, 2018, 2019, 2020, 2024): For integral-surface questions: four phases — auxiliary equations, two first integrals, general solution, surface through the curve. Each phase is ~3 marks; don’t collapse them. For orthogonal-surface questions: show the gradient computation and the resulting Lagrange PDE before solving.

Practice Set