1. Complementary Function (C.F.):
The given equation is $(D^2 + 2D + 1)y = x^2 + \sin 2x$.
The auxiliary equation is $m^2 + 2m + 1 = 0 \implies (m + 1)^2 = 0$.
Roots are real and equal: $m = -1, -1$.
Therefore, $\text{C.F.} = (C_1 + C_2 x)e^{-x}$.

2. Particular Integral 1 ($P.I._1$) for $x^2$:
$$ P.I._1 = \frac{1}{D^2 + 2D + 1} x^2 = \frac{1}{(1 + D)^2} x^2 $$
Using binomial expansion $(1+D)^{-2} = 1 - 2D + 3D^2 - 4D^3 + \dots$
$$ P.I._1 = (1 - 2D + 3D^2 - \dots) x^2 $$
Applying the operators ($D$ is derivative):
$$ P.I._1 = x^2 - 2(2x) + 3(2) = x^2 - 4x + 6 $$

3. Particular Integral 2 ($P.I._2$) for $\sin 2x$:
$$ P.I._2 = \frac{1}{D^2 + 2D + 1} \sin 2x $$
Replace $D^2$ with $-(2)^2 = -4$:
$$ P.I._2 = \frac{1}{-4 + 2D + 1} \sin 2x = \frac{1}{2D - 3} \sin 2x $$
Multiply numerator and denominator by $(2D + 3)$ to create a $D^2$ term in the denominator:
$$ P.I._2 = \frac{2D + 3}{(2D - 3)(2D + 3)} \sin 2x = \frac{2D + 3}{4D^2 - 9} \sin 2x $$
Again replace $D^2$ with $-4$:
$$ P.I._2 = \frac{2D + 3}{4(-4) - 9} \sin 2x = \frac{2D(\sin 2x) + 3\sin 2x}{-16 - 9} $$
$$ P.I._2 = \frac{2(2 \cos 2x) + 3\sin 2x}{-25} = -\frac{1}{25} (4 \cos 2x + 3 \sin 2x) $$

4. Final General Solution:
The general solution is $y = \text{C.F.} + P.I._1 + P.I._2$
$$ y = (C_1 + C_2 x)e^{-x} + x^2 - 4x + 6 - \frac{1}{25}(4 \cos 2x + 3 \sin 2x) $$

1. Complementary Function (C.F.):
Equation: $(D^2 + 1)y = \tan x$. Auxiliary equation: $m^2 + 1 = 0 \implies m = \pm i$.
$\text{C.F.} = C_1 \cos x + C_2 \sin x$.
Let $y_1 = \cos x$ and $y_2 = \sin x$.
The RHS function is $X = \tan x$.

2. Compute the Wronskian ($W$):
$$ W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x - (-\sin^2 x) = \cos^2 x + \sin^2 x = 1 $$

3. Calculate parameter $u$:
$$ u = -\int \frac{y_2 X}{W} dx = -\int \frac{\sin x \tan x}{1} dx = -\int \frac{\sin^2 x}{\cos x} dx $$
Substitute $\sin^2 x = 1 - \cos^2 x$:
$$ u = -\int \frac{1 - \cos^2 x}{\cos x} dx = -\int (\sec x - \cos x) dx $$
$$ u = -[ \ln|\sec x + \tan x| - \sin x ] = \sin x - \ln|\sec x + \tan x| $$

4. Calculate parameter $v$:
$$ v = \int \frac{y_1 X}{W} dx = \int \frac{\cos x \tan x}{1} dx = \int \cos x \left(\frac{\sin x}{\cos x}\right) dx = \int \sin x \, dx $$
$$ v = -\cos x $$

5. Final General Solution:
The general solution is $y = C_1 y_1 + C_2 y_2 + u y_1 + v y_2$.
$$ y = C_1 \cos x + C_2 \sin x + (\sin x - \ln|\sec x + \tan x|)\cos x + (-\cos x)\sin x $$
Simplify the Particular Integral part:
$P.I. = \sin x \cos x - \cos x \ln|\sec x + \tan x| - \cos x \sin x = -\cos x \ln|\sec x + \tan x|$.
$$ y = C_1 \cos x + C_2 \sin x - \cos x \ln|\sec x + \tan x| $$

1. Formula for Radius of Curvature ($\rho$):
For a polar curve $r = f(\theta)$, the radius of curvature is given by:
$$ \rho = \frac{(r^2 + r_1^2)^{3/2}}{r^2 + 2r_1^2 - r r_2} $$
Where $r_1 = \frac{dr}{d\theta}$ and $r_2 = \frac{d^2r}{d\theta^2}$.

2. Find the Derivatives:
Given curve: $r = a(1 + \cos \theta)$
First derivative ($r_1$):
$$ r_1 = \frac{d}{d\theta}[a(1 + \cos \theta)] = -a \sin \theta $$
Second derivative ($r_2$):
$$ r_2 = \frac{d}{d\theta}[-a \sin \theta] = -a \cos \theta $$

3. Evaluate at $\theta = \pi/2$:
Substitute $\theta = \pi/2$ into $r$, $r_1$, and $r_2$:
$$ r = a(1 + \cos(\pi/2)) = a(1 + 0) = a $$
$$ r_1 = -a \sin(\pi/2) = -a(1) = -a $$
$$ r_2 = -a \cos(\pi/2) = -a(0) = 0 $$

4. Calculate $\rho$:
Substitute the evaluated values into the formula:
$$ \rho = \frac{(a^2 + (-a)^2)^{3/2}}{a^2 + 2(-a)^2 - (a)(0)} $$
$$ \rho = \frac{(a^2 + a^2)^{3/2}}{a^2 + 2a^2 - 0} = \frac{(2a^2)^{3/2}}{3a^2} $$
Simplify the numerator:
$$ (2a^2)^{3/2} = 2^{3/2} \cdot (a^2)^{3/2} = 2\sqrt{2} \cdot a^3 $$
So,
$$ \rho = \frac{2\sqrt{2}a^3}{3a^2} = \frac{2\sqrt{2}a}{3} $$

1. Define Parametric Coordinates:
For the parabola $y^2 = 4ax$, the parametric equations are $x = at^2$ and $y = 2at$.

2. Find $y_1$ and $y_2$:
$$ y_1 = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} $$
$$ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dt}(y_1) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{1}{t}\right) \cdot \frac{1}{2at} = \left(-\frac{1}{t^2}\right) \frac{1}{2at} = -\frac{1}{2at^3} $$

3. Formula for Center of Curvature $(\bar{x}, \bar{y})$:
$$ \bar{x} = x - \frac{y_1(1 + y_1^2)}{y_2} $$
$$ \bar{y} = y + \frac{1 + y_1^2}{y_2} $$

4. Compute $\bar{x}$ and $\bar{y}$:
First, compute $\frac{1 + y_1^2}{y_2}$:
$$ \frac{1 + 1/t^2}{-1/(2at^3)} = -2at^3 \left(1 + \frac{1}{t^2}\right) = -2at^3 - 2at $$
Now, substitute into $\bar{x}$ and $\bar{y}$:
$$ \bar{x} = at^2 - y_1 \left( \frac{1 + y_1^2}{y_2} \right) = at^2 - \frac{1}{t}(-2at^3 - 2at) = at^2 + 2at^2 + 2a = 3at^2 + 2a $$
$$ \bar{y} = 2at + (-2at^3 - 2at) = -2at^3 $$

5. Eliminate parameter $t$ to find the Evolute:
From the $\bar{x}$ equation: $3at^2 = \bar{x} - 2a \implies t^2 = \frac{\bar{x} - 2a}{3a}$.
From the $\bar{y}$ equation: Square both sides to get $t^6$:
$$ \bar{y}^2 = (-2at^3)^2 = 4a^2 t^6 = 4a^2 (t^2)^3 $$
Substitute $t^2$ into the $\bar{y}^2$ equation:
$$ \bar{y}^2 = 4a^2 \left( \frac{\bar{x} - 2a}{3a} \right)^3 = 4a^2 \frac{(\bar{x} - 2a)^3}{27a^3} = \frac{4(\bar{x} - 2a)^3}{27a} $$
Rearrange to obtain the final equation (replacing $\bar{x}, \bar{y}$ with $x, y$):
$$ 27ay^2 = 4(x - 2a)^3 $$

1. Simplify the $n$-th term ($u_n$):
Let $u_n = \sqrt{n^3+1} - \sqrt{n^3-1}$.
To analyze its behavior as $n \to \infty$, rationalize the expression by multiplying and dividing by its conjugate:
$$ u_n = \frac{(\sqrt{n^3+1} - \sqrt{n^3-1})(\sqrt{n^3+1} + \sqrt{n^3-1})}{\sqrt{n^3+1} + \sqrt{n^3-1}} $$
Using $(a-b)(a+b) = a^2 - b^2$:
$$ u_n = \frac{(n^3+1) - (n^3-1)}{\sqrt{n^3+1} + \sqrt{n^3-1}} = \frac{2}{\sqrt{n^3+1} + \sqrt{n^3-1}} $$

2. Choose an auxiliary series ($v_n$):
For large $n$, the $+1$ and $-1$ inside the square roots become negligible.
$u_n \approx \frac{2}{\sqrt{n^3} + \sqrt{n^3}} = \frac{2}{2n^{3/2}} = \frac{1}{n^{3/2}}$.
Thus, we choose $v_n = \frac{1}{n^{3/2}}$.

3. Apply the Limit Comparison Test:
$$ \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \left( \frac{2}{\sqrt{n^3+1} + \sqrt{n^3-1}} \cdot \frac{n^{3/2}}{1} \right) $$
Divide numerator and denominator by $n^{3/2}$ (which goes inside the square root as $n^3$):
$$ = \lim_{n \to \infty} \frac{2}{\sqrt{1 + 1/n^3} + \sqrt{1 - 1/n^3}} $$
As $n \to \infty$, $1/n^3 \to 0$:
$$ = \frac{2}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2}{2} = 1 $$
Since the limit is finite and non-zero ($l \neq 0$), both $\sum u_n$ and $\sum v_n$ behave alike.

4. Conclusion (p-series test):
The auxiliary series $\sum v_n = \sum \frac{1}{n^{3/2}}$ is a p-series with $p = 3/2$.
Since $p = 3/2 > 1$, the series $\sum v_n$ is convergent.
Therefore, by the Limit Comparison Test, the given series $\sum u_n$ is also convergent.

1. Identify the general term ($u_n$):
Omitting the first term ($1$) does not alter convergence.
Let $u_n = \frac{2 \cdot 4 \cdot 6 \dots (2n)}{3 \cdot 5 \cdot 7 \dots (2n+1)}$
Then $u_{n+1} = \frac{2 \cdot 4 \cdot 6 \dots (2n)(2n+2)}{3 \cdot 5 \cdot 7 \dots (2n+1)(2n+3)}$

2. Apply D'Alembert's Ratio Test:
$$ \frac{u_n}{u_{n+1}} = \frac{2n+3}{2n+2} $$
Take the limit as $n \to \infty$:
$$ \lim_{n \to \infty} \frac{u_n}{u_{n+1}} = \lim_{n \to \infty} \frac{2 + 3/n}{2 + 2/n} = \frac{2}{2} = 1 $$
Since the limit equals 1, the Ratio Test fails. We must proceed to Raabe's Test.

3. Apply Raabe's Test:
Raabe's Test formula: $\lim_{n \to \infty} n \left( \frac{u_n}{u_{n+1}} - 1 \right)$
Calculate the expression inside the limit:
$$ \frac{u_n}{u_{n+1}} - 1 = \frac{2n+3}{2n+2} - 1 = \frac{2n+3 - (2n+2)}{2n+2} = \frac{1}{2n+2} $$
Now multiply by $n$ and take the limit:
$$ \lim_{n \to \infty} n \left( \frac{1}{2n+2} \right) = \lim_{n \to \infty} \frac{n}{2n+2} = \lim_{n \to \infty} \frac{1}{2 + 2/n} = \frac{1}{2} $$

4. Conclusion:
According to Raabe's test, if the limit $l < 1$, the series diverges. If $l > 1$, it converges.
Since $l = \frac{1}{2} < 1$, the given series is divergent.

Step 1: Write the symmetric matrix (A) of the quadratic form.
The general format is $ax^2 + by^2 + cz^2 + 2hxy + 2gxz + 2fyz$.
Comparing this, $a=3, b=-2, c=-1$. $2h = -4 \implies h=-2$. $2g = 8 \implies g=4$. $2f = 12 \implies f=6$.
$$ A = \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} = \begin{pmatrix} 3 & -2 & 4 \\ -2 & -2 & 6 \\ 4 & 6 & -1 \end{pmatrix} $$

Step 2: Find the Characteristic Equation.
The equation is $\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$.
$S_1$ (Trace) = Sum of principal diagonal elements = $3 - 2 - 1 = 0$.
$S_2$ = Sum of minors of principal diagonal elements = $\begin{vmatrix} -2 & 6 \\ 6 & -1 \end{vmatrix} + \begin{vmatrix} 3 & 4 \\ 4 & -1 \end{vmatrix} + \begin{vmatrix} 3 & -2 \\ -2 & -2 \end{vmatrix}$
$S_2 = (2 - 36) + (-3 - 16) + (-6 - 4) = -34 - 19 - 10 = -63$.
$S_3$ = Determinant $|A|$ = $3(2 - 36) - (-2)(2 - 24) + 4(-12 - 8) = 3(-34) + 2(-22) + 4(-20) = -102 - 44 - 80 = -162$.
Wait, earlier manual calculation gave $|A| = -162$. Let's re-verify $|A|$:
$3(2 - 36) = -102$. $-(-2)[(-2)(-1) - (6)(4)] = 2[2 - 24] = -44$. $4[(-2)(6) - (-2)(4)] = 4[-12 + 8] = 4[-4] = -16$.
$-102 - 44 - 16 = -162$. Correct.
The characteristic equation is $\lambda^3 - 0\lambda^2 - 63\lambda + 162 = 0$.

Step 3: Find the Eigenvalues ($\lambda$).
By inspection, $\lambda = 3$ is a root: $(3)^3 - 63(3) + 162 = 27 - 189 + 162 = 0$.
Divide by $(\lambda - 3)$ using synthetic division to get the quadratic factor: $\lambda^2 + 3\lambda - 54 = 0$.
Solving $\lambda^2 + 3\lambda - 54 = (\lambda + 9)(\lambda - 6) = 0 \implies \lambda = -9, \lambda = 6$.
The eigenvalues are $\lambda = 3, 6, -9$.

Step 4: Find the Eigenvectors.
Solve $(A - \lambda I)X = 0$.

Case 1 ($\lambda = 3$):
$$ \begin{pmatrix} 0 & -2 & 4 \\ -2 & -5 & 6 \\ 4 & 6 & -4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Using cross-multiplication on the first two rows:
$\frac{x_1}{(-2)(6) - (4)(-5)} = \frac{x_2}{(4)(-2) - (0)(6)} = \frac{x_3}{(0)(-5) - (-2)(-2)}$
$\frac{x_1}{-12 + 20} = \frac{x_2}{-8} = \frac{x_3}{-4} \implies \frac{x_1}{8} = \frac{x_2}{-8} = \frac{x_3}{-4} \implies \frac{x_1}{2} = \frac{x_2}{-2} = \frac{x_3}{-1}$.
$X_1 = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$

Case 2 ($\lambda = 6$):
$$ \begin{pmatrix} -3 & -2 & 4 \\ -2 & -8 & 6 \\ 4 & 6 & -7 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Cross-multiplication on first two rows:
$\frac{x_1}{(-2)(6) - (4)(-8)} = \frac{x_2}{(4)(-2) - (-3)(6)} = \frac{x_3}{(-3)(-8) - (-2)(-2)}$
$\frac{x_1}{-12 + 32} = \frac{x_2}{-8 + 18} = \frac{x_3}{24 - 4} \implies \frac{x_1}{20} = \frac{x_2}{10} = \frac{x_3}{20} \implies \frac{x_1}{2} = \frac{x_2}{1} = \frac{x_3}{2}$.
$X_2 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$

Case 3 ($\lambda = -9$):
$$ \begin{pmatrix} 12 & -2 & 4 \\ -2 & 7 & 6 \\ 4 & 6 & 8 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Cross-multiplication on first two rows:
$\frac{x_1}{(-2)(6) - (4)(7)} = \frac{x_2}{(4)(-2) - (12)(6)} = \frac{x_3}{(12)(7) - (-2)(-2)}$
$\frac{x_1}{-12 - 28} = \frac{x_2}{-8 - 72} = \frac{x_3}{84 - 4} \implies \frac{x_1}{-40} = \frac{x_2}{-80} = \frac{x_3}{80} \implies \frac{x_1}{1} = \frac{x_2}{2} = \frac{x_3}{-2}$.
$X_3 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$

Step 5: Form the Normalized Modal Matrix (N).
Check orthogonality: $X_1 \cdot X_2 = (2)(2)+(-2)(1)+(-1)(2) = 0$. $X_2 \cdot X_3 = 0$. $X_1 \cdot X_3 = 0$.
Find length of each eigenvector: $||X_1|| = \sqrt{2^2+(-2)^2+(-1)^2} = \sqrt{9} = 3$. Similarly, $||X_2|| = 3$ and $||X_3|| = 3$.
Divide each vector by its length to form normalized columns:
$$ N = \begin{pmatrix} 2/3 & 2/3 & 1/3 \\ -2/3 & 1/3 & 2/3 \\ -1/3 & 2/3 & -2/3 \end{pmatrix} $$ The orthogonal transformation is $X = NY$.

Step 6: Write the Canonical Form and discuss properties.
The canonical form is given by $D = N^T A N$, which is a diagonal matrix containing the eigenvalues.
Canonical Form: $\lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 \implies \mathbf{3y_1^2 + 6y_2^2 - 9y_3^2}$
- Rank (r): Number of non-zero eigenvalues = 3.
- Index (p): Number of positive eigenvalues = 2 (which are 3 and 6).
- Signature (s): $2p - r = 2(2) - 3 = 1$.
- Nature: Since there are both positive and negative eigenvalues, the nature is Indefinite.

Step 1: Set up the functions.
Let the length be $x$, width be $y$, and height be $z$.
We want to maximize the capacity (Volume), so the objective function is $V(x,y,z) = xyz$.
The box is open at the top, so it consists of 1 base ($xy$) and 4 side walls ($2xz + 2yz$).
The constraint function is Surface Area, $S = xy + 2xz + 2yz = 432$.

Step 2: Construct the Lagrangian Function.
$L(x,y,z,\lambda) = V(x,y,z) + \lambda(\text{Constraint}) = xyz + \lambda(xy + 2xz + 2yz - 432)$

Step 3: Differentiate partially w.r.t $x, y, z$ and set to zero.
$\frac{\partial L}{\partial x} = yz + \lambda(y + 2z) = 0 \implies -\lambda = \frac{yz}{y + 2z}$    --- (Eq 1)

$\frac{\partial L}{\partial y} = xz + \lambda(x + 2z) = 0 \implies -\lambda = \frac{xz}{x + 2z}$    --- (Eq 2)

$\frac{\partial L}{\partial z} = xy + \lambda(2x + 2y) = 0 \implies -\lambda = \frac{xy}{2x + 2y}$    --- (Eq 3)

Step 4: Equate the relationships to solve for $x, y, z$.
Equating (Eq 1) and (Eq 2):
$\frac{yz}{y + 2z} = \frac{xz}{x + 2z}$
Since $z \neq 0$ (otherwise volume is 0), we divide by $z$ and cross-multiply:
$y(x + 2z) = x(y + 2z) \implies xy + 2yz = xy + 2xz \implies 2yz = 2xz \implies y = x$

Equating (Eq 2) and (Eq 3):
$\frac{xz}{x + 2z} = \frac{xy}{2x + 2y}$
Since $y = x$, substitute $y$ with $x$ on the right side:
$\frac{xz}{x + 2z} = \frac{x^2}{2x + 2x} = \frac{x^2}{4x} = \frac{x}{4}$
Divide by $x$ (since $x \neq 0$):
$\frac{z}{x + 2z} = \frac{1}{4} \implies 4z = x + 2z \implies 2z = x \implies z = \frac{x}{2}$

Step 5: Apply to the Constraint Equation.
Substitute $y = x$ and $z = x/2$ into the surface area equation $xy + 2xz + 2yz = 432$:
$x(x) + 2x\left(\frac{x}{2}\right) + 2x\left(\frac{x}{2}\right) = 432$
$x^2 + x^2 + x^2 = 432$
$3x^2 = 432 \implies x^2 = 144 \implies x = 12$ cm.

Find $y$ and $z$:
$y = x \implies y = 12$ cm.
$z = \frac{x}{2} \implies z = \frac{12}{2} = 6$ cm.

Step 6: Conclusion.
The dimensions of the open rectangular box for maximum capacity are:
Length = $12$ cm, Width = $12$ cm, Height = $6$ cm.
(Maximum Volume = $12 \times 12 \times 6 = 864$ cu.cm)