← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q6a — Step-by-Step Solution

15 marks · Section B

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

Question

Find the general solution of the partial differential equation

(y3x−2x4)p+(2y4−x3y)q=9z(x3−y3),(y^3x-2x^4)p+(2y^4-x^3y)q=9z(x^3-y^3),

where p=∂z∂x, q=∂z∂yp=\dfrac{\partial z}{\partial x},\ q=\dfrac{\partial z}{\partial y}, and find its integral surface that passes through the curve x=t, y=t2, z=1x=t,\ y=t^2,\ z=1.

Technique

Lagrange auxiliary equations; clever multipliers (y,x,0)(y,x,0) and (3x2,3y2,0)(3x^2,3y^2,0) to produce exact differentials d(xy)d(xy) and d(x3+y3)d(x^3+y^3); eliminate the curve parameter tt between the two integrals.

Solution

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

P=y3x−2x4=x(y3−2x3),Q=2y4−x3y=y(2y3−x3),R=9z(x3−y3).P=y^3x-2x^4=x(y^3-2x^3),\quad Q=2y^4-x^3y=y(2y^3-x^3),\quad R=9z(x^3-y^3).

Lagrange’s auxiliary (subsidiary) equations are

dxP=dyQ=dzR.\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}.

We seek two independent first integrals u1=c1, u2=c2u_1=c_1,\ u_2=c_2; the general solution is Φ(u1,u2)=0\Phi(u_1,u_2)=0.

Step 1 — First integral via the multipliers (y,x,0)(y,x,0)

Using multipliers y, x, 0y,\,x,\,0, each fraction equals

y dx+x dyyP+xQ.\frac{y\,dx+x\,dy}{yP+xQ}.

Compute the denominator:

yP+xQ=xy(y3−2x3)+xy(2y3−x3)=xy(3y3−3x3)=−3xy(x3−y3).yP+xQ=xy(y^3-2x^3)+xy(2y^3-x^3)=xy\big(3y^3-3x^3\big)=-3xy(x^3-y^3).

The numerator is d(xy)d(xy). Equating with dzR=dz9z(x3−y3)\dfrac{dz}{R}=\dfrac{dz}{9z(x^3-y^3)}:

d(xy)−3xy(x3−y3)=dz9z(x3−y3) ⟹ d(xy)−3xy=dz9z.\frac{d(xy)}{-3xy(x^3-y^3)}=\frac{dz}{9z(x^3-y^3)}\ \Longrightarrow\ \frac{d(xy)}{-3xy}=\frac{dz}{9z}.

Integrate: −3ln⁡(xy)=ln⁡z+const-3\ln(xy)=\ln z+\text{const}, i.e.

  u1=z (xy)3=c1.  \boxed{\;u_1=z\,(xy)^3=c_1.\;}

Step 2 — Second integral via the multipliers (3x2,3y2,0)(3x^2,3y^2,0)

Using multipliers 3x2, 3y2, 03x^2,\,3y^2,\,0, each fraction equals

3x2 dx+3y2 dy3x2P+3y2Q.\frac{3x^2\,dx+3y^2\,dy}{3x^2P+3y^2Q}.

Denominator:

3x2P+3y2Q=3x3(y3−2x3)+3y3(2y3−x3)=6y6−6x6=−6(x3−y3)(x3+y3).3x^2P+3y^2Q=3x^3(y^3-2x^3)+3y^3(2y^3-x^3)=6y^6-6x^6=-6(x^3-y^3)(x^3+y^3).

Numerator is d(x3+y3)d(x^3+y^3). Equating with dz9z(x3−y3)\dfrac{dz}{9z(x^3-y^3)}:

d(x3+y3)−6(x3−y3)(x3+y3)=dz9z(x3−y3) ⟹ d(x3+y3)−6(x3+y3)=dz9z.\frac{d(x^3+y^3)}{-6(x^3-y^3)(x^3+y^3)}=\frac{dz}{9z(x^3-y^3)}\ \Longrightarrow\ \frac{d(x^3+y^3)}{-6(x^3+y^3)}=\frac{dz}{9z}.

Integrate: −3ln⁡(x3+y3)=2ln⁡z+const-3\ln(x^3+y^3)=2\ln z+\text{const}, i.e.

  u2=z2 (x3+y3)3=c2.  \boxed{\;u_2=z^2\,(x^3+y^3)^3=c_2.\;}

These are functionally independent, so the general solution is

Φ(z(xy)3, z2(x3+y3)3)=0.\Phi\big(z(xy)^3,\ z^2(x^3+y^3)^3\big)=0.

Step 3 — Integral surface through x=t, y=t2, z=1x=t,\,y=t^2,\,z=1

On the curve:

c1=z(xy)3=1⋅(t⋅t2)3=t9,c2=z2(x3+y3)3=1⋅(t3+t6)3.c_1=z(xy)^3=1\cdot(t\cdot t^2)^3=t^9,\qquad c_2=z^2(x^3+y^3)^3=1\cdot(t^3+t^6)^3.

Eliminate tt. From c1=t9c_1=t^9 we get t3=c11/3t^3=c_1^{1/3}. Then

c2=(t3+t6)3=t9(1+t3)3=c1(1+c11/3)3.c_2=(t^3+t^6)^3=t^9(1+t^3)^3=c_1\big(1+c_1^{1/3}\big)^3.

Take cube roots: c21/3=c11/3(1+c11/3)=c11/3+c12/3c_2^{1/3}=c_1^{1/3}\big(1+c_1^{1/3}\big)=c_1^{1/3}+c_1^{2/3}. Substituting back c11/3=z1/3xyc_1^{1/3}=z^{1/3}xy and c21/3=z2/3(x3+y3)c_2^{1/3}=z^{2/3}(x^3+y^3):

z2/3(x3+y3)=z1/3xy+z2/3x2y2 ⟹ z2/3(x3+y3−x2y2)=z1/3xy.z^{2/3}(x^3+y^3)=z^{1/3}xy+z^{2/3}x^2y^2\ \Longrightarrow\ z^{2/3}\big(x^3+y^3-x^2y^2\big)=z^{1/3}xy.

Cube to clear fractional powers:

Answer

  z (x3+y3−x2y2)3=(xy)3.  \boxed{\;z\,\big(x^3+y^3-x^2y^2\big)^3=(xy)^3.\;}
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