(a) Find the complete integral of z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y. (15 marks)

(b) Find the unique polynomial of degree 2 or less which fits the following data:
x : 0 1 3
f(x) : 1 3 55
Also obtain the bound on the truncation error. (15 marks)

(c) S

Question

(a) Find the complete integral of z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y. (15 marks)

(b) Find the unique polynomial of degree 2 or less which fits the following data:
x : 0 1 3
f(x) : 1 3 55
Also obtain the bound on the truncation error. (15 marks)

(c) Show that for an incompressible steady flow with constant viscosity, the velocity components
u(y) = (U/h)y - (hy/2μ)(dp/dx)(1-y/h)
v = 0 = w, with p = p(x), satisfy the equation of motion in the absence of body force. Given that U, h and dp/dx are constants. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Using Charpit’s method, take
F = z(p² − q²) − x + y = 0.
Then
F_p = 2zp, F_q = −2zq, F_x = −1, F_y = 1, F_z = p² − q².

The Charpit equations are
dx/(2zp) = dy/(−2zq) = dz/(2(x−y)) = −dp/(−1 + p(p²−q²)) = −dq/(1 + q(p²−q²)).

Since p² − q² = (x−y)/z, along a characteristic,
d(p+q)/dλ = −(p+q)(x−y)/z, dz/dλ = 2(x−y).
Thus
d(p+q)/(p+q) = −dz/(2z),
so (p+q)√z = a, where a is an arbitrary constant. Hence
p+q = a/√z.

Also
p−q = (p²−q²)/(p+q) = (x−y)/(a√z).
Therefore
p = (a² + x − y)/(2a√z),
q = (a² − x + y)/(2a√z).

Now use dz = p dx + q dy:
2a√z dz = (a² + x − y)dx + (a² − x + y)dy
= a²(dx+dy) + (x−y)(dx−dy)
= a² d(x+y) + (1/2)d((x−y)²).

Integrating,
(4a/3) z^(3/2) = a²(x+y) + (x−y)²/2 + b,
where a and b are arbitrary constants, a ≠ 0.
Complete integral: (4a/3) z^(3/2) = a²(x+y) + (x−y)²/2 + b.

(b) Using Newton’s divided-difference interpolation,
f[0] = 1,
f[0,1] = (3−1)/(1−0) = 2,
f[1,3] = (55−3)/(3−1) = 26,
f[0,1,3] = (26−2)/(3−0) = 8.

Hence
P(x) = 1 + 2x + 8x(x−1)
= 1 + 2x + 8x² − 8x
= 8x² − 6x + 1.

P(x) = 8x² − 6x + 1.

The truncation error for quadratic interpolation at nodes 0,1,3 is
E(x) = f(x) − P(x) = f'''(ξ)/6 · x(x−1)(x−3),
for some ξ between the nodes and x, provided f is three times differentiable.

On [0,3],
max |x(x−1)(x−3)| = (20 + 14√7)/27,
attained at x = (4 + √7)/3. Therefore, if
M = max_[0,3] |f'''(t)|,
then
|E(x)| ≤ M(20 + 14√7)/162 = M(10 + 7√7)/81.

Bound: |E(x)| ≤ M(10 + 7√7)/81, M = max |f'''| on the interval.

(c) For an incompressible steady flow with constant viscosity μ and no body force, the Navier–Stokes equations are
∂u/∂x + ∂v/∂y + ∂w/∂z = 0,
ρ(u∂u/∂x + v∂u/∂y + w∂u/∂z) = −∂p/∂x + μ∇²u,
and similarly for v and w.

Given
u(y) = (U/h)y − (h y/(2μ))(dp/dx)(1 − y/h),
v = 0 = w, p = p(x).

Continuity:
∂u/∂x + ∂v/∂y + ∂w/∂z = 0 + 0 + 0 = 0.
So the incompressibility condition is satisfied.

Let P = dp/dx. Then
u = (U/h)y − (hP/(2μ))y + (P/(2μ))y².
Differentiate:
du/dy = U/h − hP/(2μ) + (P/μ)y,
d²u/dy² = P/μ.

For the x-momentum equation, since u = u(y), v = w = 0,
ρ(u∂u/∂x + v∂u/∂y + w∂u/∂z) = 0.
Also ∂²u/∂x² = ∂²u/∂z² = 0, so
−∂p/∂x + μ d²u/dy² = −P + μ(P/μ) = 0.
Thus the x-momentum equation is satisfied.

For the y-momentum equation, v = 0 and p = p(x) imply
−∂p/∂y + μ∇²v = 0 + 0 = 0.
Similarly, the z-momentum equation is satisfied.

Hence the given velocity field and pressure satisfy the equation of motion in the absence of body force.

(a) Find the complete integral of z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y. (15 marks) (b) Find the unique polynomial of degree 2 or…

Model answer

What "Solve" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Mathematics 2025 Paper II