Charpit’s method

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2016, 2017, 2025)
• Priority tier: T3
• Marks (count): 15 (3)
• Average solve time: ~16 min
• Difficulty mix: hard 2, medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

Charpit’s method is the standard tool for nonlinear first-order PDEs in UPSC Paper 2, appearing with a 15-mark allocation in 2016, 2017, and 2025. The technique has a fixed pipeline but demands algebraic skill at two points: identifying the clean first integral from the auxiliary ratios, and executing the final integration of $dz=p\,dx+q\,dy$. The 2016 question (characteristic-strip on a parabola) and the 2025 question (separable structure leading to $z^{3/2}$) illustrate that the method is versatile — knowing it covers both complete-integral and integral-surface variants.

Minimum Theory

Charpit’s auxiliary equations. For a nonlinear first-order PDE $F(x,y,z,p,q)=0$ ($p=z_x,q=z_y$), Charpit’s method seeks a one-parameter family compatible with $F=0$. The auxiliary (Charpit) equations are:

$\frac{dp}{F_x+pF_z}=\frac{dq}{F_y+qF_z}=\frac{dz}{-(pF_p+qF_q)}=\frac{dx}{-F_p}=\frac{dy}{-F_q}.$

Any first integral $f(x,y,z,p,q)=a$ of this system, combined with $F=0$, gives compatible equations $p=p(x,y,z,a)$, $q=q(x,y,z,a)$. Integration of $dz=p\,dx+q\,dy$ then yields the complete integral $z=\phi(x,y,a,b)$ containing two arbitrary constants $a,b$.

Finding the first integral. The most productive ratio pairs are those in which the numerators are each other’s negatives after algebra, giving $d(\text{combination})=0$. Common patterns: $dp+dx$ (yielding $p+x=a$), $d(p+x)+d(q+y)=0$ (yielding $(p+x)+(q+y)=a$), or separable structure $F=f(x,p,z)-g(y,q,z)$ (yielding $f=g=a$).

Characteristic-strip for complete integral. When the question asks to find the integral surface through an initial curve, the same Charpit equations are used as characteristic-strip equations; $p_0,q_0$ on the initial curve are determined by the PDE and the strip condition $\dot z_0=p_0\dot x_0+q_0\dot y_0$.

Question Archetypes

charpit-method (2 question(s); 2017, 2025)

Recognition Cues — The question asks for a “complete integral” of a nonlinear first-order PDE. In 2017 the PDE factors via a substitution $P=p+x$, $Q=q+y$; in 2025 the PDE separates as $zp^2-x=zq^2-y$.

Solution Template

1. Write $F(x,y,z,p,q)=0$ and compute $F_x,F_y,F_z,F_p,F_q$.
2. Write Charpit’s five-ratio chain.
3. Identify the first integral: add ratio numerators to form an exact differential; or spot a separable $x$-only vs.\ $y$-only structure in $F$.
4. Combine the first integral with $F=0$ to express $p$ and $q$ explicitly in terms of $x,y,z,a$.
5. Integrate $dz=p\,dx+q\,dy$ (check it is exact: $\partial p/\partial y=\partial q/\partial x$).
6. Write the complete integral $z=\phi(x,y,a,b)$ with two constants; check by substituting back.

Worked Examples

2017 Paper 2, 2017-P2-Q6a (15 marks)

Find a complete integral of $2(pq+yp+qx)+x^2+y^2=0$.

Write $F=2pq+2yp+2qx+x^2+y^2=0$. Note $F_z=0$. Partials:

$F_x=2q+2x,\quad F_y=2p+2y,\quad F_p=2q+2y,\quad F_q=2p+2x.$

Charpit ratios ($F_z=0$):

$\frac{dp}{2(q+x)}=\frac{dq}{2(p+y)}=\frac{dx}{-2(q+y)}=\frac{dy}{-2(p+x)}.$

From the $p$- and $x$-ratios: $dp+dx$ corresponds to numerator $(2(q+x))+(- 2(q+y))=2(x-y)$ and from the $q$- and $y$-ratios: $dq+dy$ corresponds to $2(p+y)+(-2(p+x))=2(y-x)$. So $d(p+x)+d(q+y)=0$, giving

$\boxed{(p+x)+(q+y)=a.}$

Rewrite the PDE. Set $P=p+x$, $Q=q+y$. Then $PQ=(p+x)(q+y)=pq+py+xq+xy$, so

$F=2PQ-2xy+x^2+y^2=2PQ+(x-y)^2=0\implies PQ=-\tfrac12(x-y)^2.$

With $P+Q=a$ and $PQ=-\tfrac12(x-y)^2$, the values $P,Q$ are roots of $t^2-at-\tfrac12(x-y)^2=0$:

$P=\frac{a+\sqrt{a^2+2(x-y)^2}}{2},\quad Q=\frac{a-\sqrt{a^2+2(x-y)^2}}{2}.$

So $p=P-x$, $q=Q-y$. Integrate $dz=p\,dx+q\,dy=(P-x)dx+(Q-y)dy$:

$dz=\frac{a}{2}(dx+dy)-(x\,dx+y\,dy)+\frac{\sqrt{a^2+2(x-y)^2}}{2}(dx-dy).$

Let $s=x-y$; the radical term is $\frac12\sqrt{a^2+2s^2}\,ds$, integrating to $\frac{s}{4}\sqrt{a^2+2s^2}+\frac{a^2}{4\sqrt2}\sinh^{-1}\!\frac{\sqrt2\,s}{a}$.

$\boxed{z=\frac{a}{2}(x+y)-\frac{x^2+y^2}{2}+\frac{x-y}{4}\sqrt{a^2+2(x-y)^2}+\frac{a^2}{4\sqrt{2}}\sinh^{-1}\!\frac{\sqrt{2}\,(x-y)}{a}+b}$

Two arbitrary constants $a,b$ confirm this is a complete integral.

2025 Paper 2, 2025-P2-Q7a (15 marks)

Find the complete integral of $z(p^2-q^2)=x-y$.

Write $F=zp^2-zq^2-x+y=0$. Partials:

$F_x=-1,\quad F_y=1,\quad F_z=p^2-q^2,\quad F_p=2zp,\quad F_q=-2zq.$

Separable structure. Rewrite the PDE as

$zp^2-x=zq^2-y.$

The left side depends only on $(x,p,z)$ and the right side only on $(y,q,z)$; for this equality to hold for all $x,y$ both sides must be a constant, say $a$:

$zp^2-x=a\implies p=\sqrt{\frac{x+a}{z}},\quad zq^2-y=a\implies q=\sqrt{\frac{y+a}{z}}.$

Check in the PDE: $z(p^2-q^2)=z\cdot\frac{x+a}{z}-z\cdot\frac{y+a}{z}=(x+a)-(y+a)=x-y$ ✓.

Integrate $dz=p\,dx+q\,dy$:

$dz=\sqrt{\frac{x+a}{z}}\,dx+\sqrt{\frac{y+a}{z}}\,dy\implies \sqrt{z}\,dz=\sqrt{x+a}\,dx+\sqrt{y+a}\,dy.$

Integrating both sides:

$\frac{2}{3}z^{3/2}=\frac{2}{3}(x+a)^{3/2}+\frac{2}{3}(y+a)^{3/2}+\frac{2}{3}b.$

$\boxed{z^{3/2}=(x+a)^{3/2}+(y+a)^{3/2}+b}$

Two arbitrary constants $a,b$: a genuine complete integral.

Common Traps

• The decisive regrouping (2017). Without spotting that $F=2(p+x)(q+y)+(x-y)^2$, the Charpit ratios look intractable. The step $d(p+x)+d(q+y)=0$ follows from noting that the sum of the $dp$- and $dx$-numerators equals the negative of the sum of the $dq$- and $dy$-numerators.
• The $\sinh^{-1}$ in 2017. The integral $\int\sqrt{a^2+2s^2}\,ds$ gives a $\sinh^{-1}$ (equivalently $\ln$) term — this is correct, not an error. The form $dz=p\,dx+q\,dy$ is exact ($p_y=q_x$ can be verified), so the line integral is path-independent and the integration is legitimate.
• Separable split in 2025. The key insight is that $zp^2-x$ is a function of $(x,p,z)$ alone and $zq^2-y$ of $(y,q,z)$ alone, so both equal a constant $a$. Don’t use the full Charpit ratio machinery when this structure is visible — it leads to the same answer faster.
• The $\sqrt{z}\,dz$ integration. In 2025, $\int\sqrt{z}\,dz=\frac{2}{3}z^{3/2}$; the final form $z^{3/2}$ should be left as is — expanding or solving for $z$ obscures the clean two-parameter structure.
• Two constants are mandatory. A “complete integral” must contain exactly two independent arbitrary constants. Forgetting $b$ (the constant of integration from $dz$) gives only a one-parameter family, which is not complete.

characteristic-strip (1 question(s); 2016)

Recognition Cues — “Determine the characteristics of $[F=0]$” paired with “find the integral surface through [curve]”. The initial curve is given as an equation in $x,y,z$ (here a parabola).

Solution Template

1. Write $F=0$; compute $F_p,F_q,F_x,F_y,F_z$.
2. The strip equations: $\dot x=F_p$, $\dot y=F_q$, $\dot z=p\dot x+q\dot y$, $\dot p=-(F_x+pF_z)$, $\dot q=-(F_y+qF_z)$ (dot $=d/ds$).
3. Integrate from initial values $(x_0,y_0,z_0,p_0,q_0)$.
4. Find $p_0,q_0$ on the initial curve: (a) the PDE, (b) the strip condition $\dot z_0=p_0\dot x_0+q_0\dot y_0$.
5. Substitute; eliminate the two parameters to express $z$ in terms of $x,y$.

Worked Example

2016 Paper 2, 2016-P2-Q6a (15 marks)

Determine the characteristics of $z=p^2-q^2$ and find the integral surface through $4z+x^2=0$, $y=0$.

Write $F=p^2-q^2-z=0$. Partials: $F_p=2p$, $F_q=-2q$, $F_z=-1$, $F_x=F_y=0$.

Strip equations:

$\dot x=2p,\quad \dot y=-2q,\quad \dot z=2p^2-2q^2=2z,\quad \dot p=p,\quad \dot q=q.$

Integrate: $p=p_0 e^s$, $q=q_0 e^s$, $z=z_0 e^{2s}$,

$x=x_0+2p_0(e^s-1),\quad y=y_0-2q_0(e^s-1).$

The characteristics are parametric curves $(x_0+2p_0(e^s-1),\;y_0-2q_0(e^s-1),\;z_0 e^{2s})$; their projections onto the $xy$-plane are straight lines with direction $(p_0,-q_0)$.

Initial data on the parabola $4z+x^2=0$, $y=0$. Parametrise: $x_0=t$, $y_0=0$, $z_0=-t^2/4$.

Strip condition on $y=0$ (so $dy_0=0$): $\dot z_0=p_0\dot x_0 \Rightarrow -t/2=p_0\cdot1$, so $p_0=-t/2$.

PDE: $p_0^2-q_0^2=z_0 \Rightarrow t^2/4-q_0^2=-t^2/4 \Rightarrow q_0^2=t^2/2$, so $q_0=t/\sqrt{2}$.

Integrate and eliminate. With these initial values:

$x=t(2-e^s),\quad y=-\sqrt{2}\,t(e^s-1),\quad z=-\tfrac{t^2}{4}e^{2s}.$

From $y=-\sqrt{2}\,t(e^s-1)$: $te^s=t+y/\sqrt{2}\cdot(-1)$… simplifying: note $x=t(2-e^s)=2t-te^s$ and $y=\sqrt{2}\,t(1-e^s)=-\sqrt{2}(te^s-t)$, so $te^s=t-y/\sqrt{2}$ and $x=2t-(t-y/\sqrt{2})=t+y/\sqrt{2}$.

Thus $t=x-y/\sqrt{2}$ and $te^s=x-y/\sqrt{2}-(y/\sqrt{2})=x-\sqrt{2}\,y$. Then

$z=-\tfrac{(te^s)^2}{4}=-\tfrac{(x-\sqrt{2}\,y)^2}{4}.$

$\boxed{z=-\frac{1}{4}(x-\sqrt{2}\,y)^2,\quad\text{i.e.}\quad 4z+(x-\sqrt{2}\,y)^2=0}$

This is a parabolic cylinder; at $y=0$, $4z+x^2=0$ ✓.

Common Traps

• Two conditions for $p_0,q_0$. Both the PDE and the strip condition are needed. On $y=0$ the strip condition reduces to $p_0=dz_0/dx_0=-t/2$; then the PDE gives $q_0^2=p_0^2-z_0=t^2/4+t^2/4=t^2/2$. Forgetting one condition leaves the system underdetermined.
• The $\dot z=2z$ consistency check. Verify early that $p^2-q^2=(p_0^2-q_0^2)e^{2s}=z_0 e^{2s}=z$ — this confirms the $z$-equation and catches arithmetic errors before the elimination step.
• The final surface is a perfect square. The answer $-\tfrac14(x-\sqrt2\,y)^2$ is a parabolic cylinder, not a cone; the $(x-\sqrt2\,y)$ combination comes naturally from eliminating $t,s$.

Marks-Aware Writing

15-mark questions (all three appearances). A full-marks response must show: (1) $F$ and all five partial derivatives written explicitly; (2) the Charpit ratio chain; (3) the first integral or separable split derived with algebra visible; (4) the expressions for $p$ and $q$ substituted into $dz=p\,dx+q\,dy$; (5) the integration (including the $\sinh^{-1}$ in 2017 or the $\frac{2}{3}z^{3/2}$ in 2025); and (6) the complete integral boxed with the statement “two arbitrary constants $a,b$”.

For the characteristic-strip question (2016), additionally state the nature of the characteristics (“parametric curves …, projecting as straight lines”) before moving to the initial data step.

Practice Set