← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 9× in 13 yrs · Read the full method →

Question

Find the general integral of the partial differential equation (y+zx)p−(x+yz)q=x2−y2(y+zx)p-(x+yz)q=x^2-y^2.

Technique

Lagrange’s method — auxiliary equations dxP=dyQ=dzR\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}; two clever multiplier sets: (x,y,z)(x,y,z) gives xP+yQ+zR=zRxP+yQ+zR=zR leading to x2+y2−z2=c1x^2+y^2-z^2=c_1; (y,x,1)(y,x,1) gives yP+xQ=−RyP+xQ=-R leading to xy+z=c2xy+z=c_2.

Solution

Setup. This is Lagrange’s linear PDE Pp+Qq=RPp+Qq=R with

P=y+zx,Q=−(x+yz),R=x2−y2.P=y+zx,\qquad Q=-(x+yz),\qquad R=x^2-y^2.

The general integral is F(u1,u2)=0F(u_1,u_2)=0, where u1=c1, u2=c2u_1=c_1,\ u_2=c_2 are two independent first integrals of the auxiliary (subsidiary) equations

dxy+zx=dy−(x+yz)=dzx2−y2.\frac{dx}{y+zx}=\frac{dy}{-(x+yz)}=\frac{dz}{x^2-y^2}.

Step 1 — First integral via the multiplier (x,y,−z)(x,y,-z)… (find x2+y2−z2x^2+y^2-z^2)

Use multipliers (x,y,z)(x,y,z) on the numerators/denominators:

x P+y Q+z R=x(y+zx)−y(x+yz)+z(x2−y2)=x2z−y2z=z(x2−y2)=z R.x\,P+y\,Q+z\,R=x(y+zx)-y(x+yz)+z(x^2-y^2)=x^2z-y^2z=z(x^2-y^2)=z\,R.

Hence

x dx+y dy+z dzz(x2−y2)=dzx2−y2 ⟹ x dx+y dy+z dz=z dz.\frac{x\,dx+y\,dy+z\,dz}{z(x^2-y^2)}=\frac{dz}{x^2-y^2}\ \Longrightarrow\ x\,dx+y\,dy+z\,dz=z\,dz.

That is x dx+y dy−z dz=0x\,dx+y\,dy-z\,dz=0, integrating to

u1=x2+y2−z2=c1.\boxed{u_1=x^2+y^2-z^2=c_1.}

Step 2 — Second integral via the multiplier (y,x,1)(y,x,1) (find xy+zxy+z)

Use multipliers (y,x,1)(y,x,1):

y P+x Q=y(y+zx)−x(x+yz)=y2−x2=−(x2−y2)=−R.y\,P+x\,Q=y(y+zx)-x(x+yz)=y^2-x^2=-(x^2-y^2)=-R.

So

y dx+x dy−R=dzR ⟹ y dx+x dy=−dz ⟹ d(xy)+dz=0.\frac{y\,dx+x\,dy}{-R}=\frac{dz}{R}\ \Longrightarrow\ y\,dx+x\,dy=-dz\ \Longrightarrow\ d(xy)+dz=0.

Integrating,

u2=xy+z=c2.\boxed{u_2=xy+z=c_2.}

Step 3 — General integral

Answer

  F(x2+y2−z2,  xy+z)=0,  \boxed{\;F\big(x^2+y^2-z^2,\ \ xy+z\big)=0,\;}
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