← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q6a — Step-by-Step Solution

15 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 10× in 14 yrs · Read the full method →

Question

Solve for the general solution $p\cos(x+y)+q\sin(x+y)=z$, where $p=\partial z/\partial x$ and $q=\partial z/\partial y$.

Technique

Lagrange auxiliary system; clever substitution $u=x+y$, $v=x-y$ to decouple; first integral via $du/(cos u+\sin u)=dz/z$ (giving the half-angle tangent form); second integral via $du/(\cos u+\sin u)=dv/(\cos u-\sin u)$ (giving $v-\ln|\cos u+\sin u|$).

Solution

Lagrange’s method. Auxiliary system:

Step 1 — First integral

Add the first two ratios:

Let $u=x+y$, $du=dx+dy$:

The LHS denominator: $\cos u+\sin u=\sqrt 2\sin(u+\pi/4)$ (or equivalently $\sqrt 2\cos(u-\pi/4)$).

So

Integrate. LHS: $\dfrac{1}{\sqrt 2}\int\csc(u+\pi/4)\,du=\dfrac{1}{\sqrt 2}\ln|\tan((u+\pi/4)/2)|+\text{const}$.

(Using $\int\csc v\,dv=\ln|\tan(v/2)|$.)

RHS: $\ln|z|+\text{const}$.

So

Exponentiate: $\left|\tan\!\left(\dfrac{x+y}{2}+\dfrac{\pi}{8}\right)\right|^{1/\sqrt 2}=A_1\cdot|z|$, i.e.

Step 2 — Second integral

Subtract the first two ratios:

Note $\cos u-\sin u=\sqrt 2\cos(u+\pi/4)$.

Let $w=x-y$ (independent variable), then $x+y=u$ is independent. We have $dw$ in the numerator, $\sqrt 2\cos(u+\pi/4)$ in the denominator (depends on $u$, not $w$). Hmm — these aren’t immediately separable.

Different approach: integrate $dx-dy$ and $dz/z$ directly?

$\dfrac{d(x-y)}{\cos(x+y)-\sin(x+y)}=\dfrac{dz}{z}$.

To proceed, eliminate using the relation from Step 1: $u=x+y$ is implicitly tied to $z$ via $(\star)$. Then $\cos u-\sin u$ can be expressed in terms of $z$, and we have $d(x-y)$ in terms of $z$ and the differential.

Alternative: use the first integral $(\star)$ to express $u$ in terms of $z$, then $\cos u-\sin u$ becomes a function of $z$ alone. Integrating $d(x-y)=f(z)\,dz/z$ gives $x-y$ as a function of $z$, a second integral.

This is getting algebraically complex. Let me try a cleaner approach.

Step 2’ — Alternative second integral via the differential of $u-2y$ or similar

Try: $\dfrac{dz}{z}=\dfrac{dx}{\cos u}=\dfrac{dy}{\sin u}$. So $dz/z$ along the characteristics has two expressions; equate them?

Easier: subtract again. $\dfrac{dx}{\cos u}-\dfrac{dy}{\sin u}=\dfrac{\sin u\,dx-\cos u\,dy}{\sin u\cos u}$. Set this equal to some other ratio?

Let me try $\dfrac{\sin u\,dx-\cos u\,dy}{\sin u\cos u-\sin u\cos u}=\dfrac{...}{0}$ — degenerate.

Cleaner: use multipliers $\sin u, -\cos u$:

So $\sin u\,dx-\cos u\,dy$ must equal 0 along the characteristics. Let’s check: $\sin u\,dx-\cos u\,dy$: along $dx/\cos u=dy/\sin u=k\,dt$ (parameterise), $dx=k\cos u\,dt$, $dy=k\sin u\,dt$. Then $\sin u\,dx-\cos u\,dy=k(\sin u\cos u-\cos u\sin u)dt=0$ ✓.

So $\sin u\,dx-\cos u\,dy=0$ along characteristics. But $u=x+y$ depends on $(x,y)$, so this isn’t an exact differential immediately. Let’s try:

$d(\sin u\,x-\cos u\,y\cdot\text{stuff})$… let me try $d(x\sin u+y\cos u)$: $d(x\sin u+y\cos u)=\sin u\,dx+x\cos u\,du+\cos u\,dy-y\sin u\,du=(\sin u\,dx+\cos u\,dy)+(x\cos u-y\sin u)du$.

Hmm — not matching $\sin u\,dx-\cos u\,dy$.

Try $d(x\sin u-y\cos u)$? $=\sin u\,dx+x\cos u\,du-\cos u\,dy+y\sin u\,du=(\sin u\,dx-\cos u\,dy)+(x\cos u+y\sin u)du$.

Closer but extra term $(x\cos u+y\sin u)du\ne 0$ along characteristics.

This problem is genuinely awkward. Let me try yet another approach.

Step 2” — Multipliers $(1,1,0)$ then $(1,-1,0)$

$(1,1,0)$: numerator $dx+dy$, denom $\cos u+\sin u$. Already used (Step 1).

$(1,-1,0)$: numerator $dx-dy$, denom $\cos u-\sin u$.

So $\dfrac{dx-dy}{\cos u-\sin u}=\dfrac{dz}{z}$ (using the third ratio).

Wait — that uses the property that if $A/a=B/b=C/c$, then for any $\lambda,\mu$: $(\lambda A+\mu B)/(\lambda a+\mu b)=C/c$ (provided denom nonzero). So $\dfrac{dx-dy}{\cos u-\sin u}=\dfrac{dz}{z}$.

Now $d(x-y)=dx-dy$, but the denom $\cos u-\sin u=\cos(x+y)-\sin(x+y)$ depends on $(x+y)$. So integration isn’t straightforward unless we change variable.

Hmm — note that we already have Step 1’s relation expressing $u$ in terms of $z$ (and $C_1$). So along a characteristic, $u$ is a function of $z$, and $\cos u-\sin u$ becomes a function of $z$:

$\cos u-\sin u=\sqrt 2\cos(u+\pi/4)$.

From Step 1: $\tan(u/2+\pi/8)=B z^{\sqrt 2}$ where $B$ depends on $C_1$. The connection from $u$ to $z$ is implicit.

This is getting complicated. Let me try the symmetric approach.

Step 2''' — Use $dx-dy$ and $dy/(\sin u)$

Try $\dfrac{dx-dy}{\cos u-\sin u}=\dfrac{dz}{z}$ — this gives a relation between $d(x-y)$ and $dz$.

But we also have $u=x+y$, and from Step 1, $u$ relates to $z$ via $\tan(u/2+\pi/8)=$ const $\cdot z^{\sqrt 2}$.

So $\cos u-\sin u$ is a function of $z$. Let me denote $F(z)=\cos u(z)-\sin u(z)$, where $u(z)$ is the function defined by the Step 1 first integral $\tan(u/2+\pi/8)=Cz^{\sqrt 2}$.

Then $d(x-y)=F(z)\dfrac{dz}{z}$, and integrating gives $x-y=G(z)+C_2$ for some $G$. The second integral is $x-y-G(z)=C_2$.

This works in principle but the explicit $G(z)$ is messy.

Simpler approach: parametric characteristics

Maybe the cleanest answer is to leave the general solution in terms of two arbitrary integrals:

with the second integral implicitly defined by the relation derived above.

A different formulation: $p\cos u+q\sin u=z$ where $u=x+y$

Since the coefficient is a function of $u=x+y$ only, substitute $u=x+y$, $v=x-y$ as new independent variables. Then $\partial/\partial x=\partial/\partial u+\partial/\partial v$, $\partial/\partial y=\partial/\partial u-\partial/\partial v$.

$p=z_x=z_u+z_v$, $q=z_y=z_u-z_v$.

PDE: $(z_u+z_v)\cos u+(z_u-z_v)\sin u=z$, i.e.

This is still a first-order linear PDE in $(u,v)$ but with $u$-dependent coefficients.

Characteristics: $\dfrac{du}{\cos u+\sin u}=\dfrac{dv}{\cos u-\sin u}=\dfrac{dz}{z}$.

First integral (from $du,dz$): as in Step 1, $\dfrac{[\tan(u/2+\pi/8)]^{1/\sqrt 2}}{z}=C_1$.

Second integral (from $du,dv$): $\dfrac{dv}{du}=\dfrac{\cos u-\sin u}{\cos u+\sin u}$. RHS depends only on $u$, so integrable:

$\dfrac{\cos u-\sin u}{\cos u+\sin u}=\dfrac{d}{du}\ln|\cos u+\sin u|=\dfrac{-\sin u+\cos u}{\sin u+\cos u}$ ✓. So

Step 3 — General solution

Answer