Applied Mathematics and Statistical Methods Solutions
1. Partial Differential Equations and Probability
(a) Solve the PDE (D² − 3DDʹ)z = 0
The auxiliary equation is obtained by replacing D with m and Dʹ with 1:
m² − 3m = 0 ⇒ m(m − 3) = 0 ⇒ m = 0, 3
Solution: z = f₁(y) + f₂(y + 3x)
(b) Classify the PDE: 4∂²u/∂x² − 4∂²u/∂x∂t + ∂²u/∂t² = 0
Comparing with A·u_xx + B·u_xt + C·u_tt = 0: A = 4, B = −4, C = 1
Discriminant = B² − 4AC = 16 − 16 = 0
Since B² − 4AC = 0, the equation is Parabolic.
(c) Rank Correlation Formula for Tied Ranks
ρ = 1 − 6[Σd² + Σ(m³ − m)/12] / [n(n² − 1)]
- d = difference between ranks of paired values
- m = number of items having the same rank in a tied group
- n = number of pairs
(d) Probability Density Function
For a valid PDF, the total area under the curve must equal 1:
∫ₐᵇ (k/x) dx = 1 ⇒ k · [ln x]ₐᵇ = 1 ⇒ k · ln(b/a) = 1
k = 1 / ln(b/a)
(e) Probability of At Least One Head
Sample space S = {HH, HT, TH, TT}, n(S) = 4
Favourable cases = {HH, HT, TH}, n(F) = 3
P(at least one head) = 3/4 = 0.75
(f) Mean of Binomial Distribution
For (1/3 + 2/3)¹⁰, where q = 1/3, p = 2/3, n = 10:
Mean = np = 10 × (2/3) = 20/3 ≈ 6.67
(g) Control Limits of C-Chart
Let c̄ be the average number of defects:
- UCL = c̄ + 3√c̄
- CL = c̄
- LCL = c̄ − 3√c̄ (if negative, take LCL = 0)
2. Advanced PDE and Regression
(a) Poisson’s Equation Solution
Given ∇²V = −4π(x² + y²), the particular solution is V_p = −(π/4)(x² + y²)².
Adding the harmonic function V_h = (π/4)(x⁴ − 6x²y² + y⁴) to satisfy V(x, 0) = 0:
V = −2πx²y²
(b) Wave Equation for Stretched String
Using the identity sin³θ = (3 sinθ − sin 3θ)/4, the displacement is:
y(x, t) = (3y₀/4) sin(πx/l) cos(πct/l) − (y₀/4) sin(3πx/l) cos(3πct/l)
(c) Lines of Regression
For the given data, the regression lines are:
- y on x: y = 0.65x + 4.10
- x on y: x = 1.30y − 4.40
- Correlation coefficient (r): ≈ 0.919
(d) Poisson as a Limiting Case of Binomial
As n → ∞ and p → 0 such that np = λ, the Binomial distribution P(X = r) converges to (e⁻ᵏ · λʳ) / r!, which is the Poisson distribution.
(e) np-chart vs p-chart
The np-chart plots the number of defectives (constant sample size), while the p-chart plots the proportion of defectives (variable sample size). For the provided data, the process is in statistical control as all values lie within (0.41, 17.59).
3. Fourier Transforms and Heat Equation
(a) Fourier Transform of F(x) = 1 − x²
F(ω) = 4[sin(ω) − ω cos(ω)] / (ω³ √(2π))
(b) Heat Equation in an Insulated Rod
The temperature distribution is given by the series:
u(x, t) = (200/π) Σ_{n=1}^∞ [(−1)^(n+1) / n] · sin(nπx/l) · exp(−c²n²π²t/l²)
4. Statistical Analysis
(a) Least Square Fit
The best-fit curve for the data is f(x) = 1.2 + 1.1x + 1.5x².
(b) Skewness and Kurtosis
Calculated coefficients indicate the distribution is slightly positively skewed (γ₁ ≈ 0.011) and platykurtic (β₂ ≈ 2.048).
(c) Chi-Square Test for Binomial Law
Calculated χ² = 19.63. Since 19.63 > 9.488 (critical value at 4 df), the data is not consistent with the hypothesis.
(d) Normal Distribution Analysis
For μ = 750 and σ = 50, the lowest income among the richest 100 persons is Rs. 866.50.
(e) T-Test and Association
The t-test (t = 0.155) indicates sailors are not significantly taller than soldiers. The Chi-square test (χ² = 16.04) confirms a significant association between Area and Pollution Index.