Engineering Mathematics Test 2

Concept:

Laurentz Series is obtained by the arrangement and manipulation of standard series or expansions, i.e.

(1 - x)-1 = 1 + x + x2 + x3 + …… |x| < 1

(1 + x)-1 = 1 – x + x2 – x3 + ….. |x| < 1

(1 - x)-2 = 1 + 2x + 3x2 + ….. |x| < 1

(1 + x)-2 1 – 2x + 3x2 – 4x2 + …. |x| < 1

Observe that in all the expansions; |x| should be less than 1.

 ∴ We need to manipulate the variable to satisfy the above condition.

Calculation:

Given:

\(f\left( z \right) = \frac{1}{{\left( {z - 1} \right)\left( {z - 2} \right)}}\)

\(= \frac{1}{{z - 2}} - \frac{1}{{z - 1}}\)

\(= \frac{1}{{\left( {z - 1 - 1} \right)}} - \frac{1}{{\left( {z - 1} \right)}}\)

\(= \frac{{ - 1}}{{\left[ {1 - \left( {z - 1} \right)} \right]}} - \frac{1}{{\left( {z - 1} \right)}}\)

= -1 [1 - (z - 1)]^-1 - [z - 1]^-1

= -[z - 1]^-1 - [1 + (z - 1) + (z - 1)² + (z - 1)³ +…]

The complex derivative of a function f at a point z₀ is defined as the limit

\(f'\left( {{z_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{z_0} + h} \right) - f\left( {{z_0}} \right)}}{h}\) whenever the limit exists.

f(z) = |z|²

At z = 0,

\(\mathop {\lim }\limits_{h \to 0} \frac{{{{\left| h \right|}^2}}}{h} = \mathop {\lim }\limits_{\left( {{h_1},{h_2}} \right) \to 0} \frac{{h_1^2 + h_2^2}}{{{h_1} + i{h_2}}} = \mathop {\lim }\limits_{\left( {{h_1},{h_2}} \right) \to 0} \frac{{h_1^2 + h_2^2}}{{h_1^2 + h_2^2}}\left( {{h_1} - i{h_2}} \right) = 0\)

(We have used the important fact that |z|² = zz̅)

Hence f(z) is differentiable at z = 0.

At z ≠ 0,

\(\mathop {\lim }\limits_{h \to 0} \frac{{{{\left| {z + h} \right|}^2} - {{\left| z \right|}^2}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{z\bar h + \bar zh}}{h}\)

The above limit does not exist for z ≠ 0 as \(\mathop {\lim }\limits_{h \to 0} \frac{{\bar h}}{h}\) does not exist.

Hence f(z) is not differentiable at z ≠ 0.

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A

The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

Calculation:

Characteristic equation: |A – λI| = 0

\(\left[ {A - \lambda I} \right] = \left[ {\begin{array}{*{20}{c}} x&0&0\\ 0&y&{ - 1}\\ 0&1&{ - 2} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} \lambda &0&0\\ 0&\lambda &0\\ 0&0&\lambda \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {x - \lambda }&0&0\\ 0&{y - \lambda }&{ - 1}\\ 0&1&{ - 2 - \lambda } \end{array}} \right]\)

\(\left| {\begin{array}{*{20}{c}} {x - \lambda }&0&0\\ 0&{\left( {y - \lambda } \right)}&{ - 1}\\ 0&1&{ - 2 - \lambda } \end{array}} \right| = 0\)

⇒ (x - λ) [(y - λ)(-λ - 2) + 1] = 0

⇒ (λ - x) [(λ - y)(λ + 2) + 1] = 0

To get all the negative Eigen values

x < 0 and the roots of [(λ - y)(λ + 2) + 1] must be negative.

(λ - y)(λ + 2) + 1 = 0

λ² – yλ + 2λ – 2y + 1 = 0

⇒ λ² + λ(2 - y) + (1 – 2y) = 0

To get the negative roots,

2 – y > 0 and 1 – 2y > 0

\( \Rightarrow y < 2\;and\;y < \frac{1}{2}\)

\( \Rightarrow y < \frac{1}{2}\)

Eigen values of B = 0, 1, 2

Determinant of B = product of Eigen values = 0

As determinant is zero, the rank of the matrix B is less than or equal to 2.

As there are two unique Eigen values other than zero, the rank of matrix B is 2.

From the properties of determinant,

Det (B^T B) = Det (B^T) ⋅ Det (B) = 0

Eigen values of B² = 0, 1², 2² = 0, 1, 4

Eigen values of B² + I = 1, 2, 5

Eigen values of (B² + I)^-1 \(= 1,\;\frac{1}{2},\frac{1}{5}\)

We can’t find the Eigen values of B^T B with the given information.

f(x) = x

f(x) is an odd function.

So, only b_n exists in the Fourier series expansion.

For -π ≤ x ≤ π,

\({b_n} = \frac{2}{\pi }\mathop \smallint \limits_0^\pi x\sin \left( {\frac{{n\pi }}{\pi }x} \right)dx\) 

\( = \frac{2}{\pi }\mathop \smallint \limits_0^\pi x\sin nx\;\;dx\) 

\( = \frac{2}{\pi }\left[ {\left[ {\frac{x}{n}\left( { - \cos nx} \right)} \right]_0^\pi - \frac{1}{n}\mathop \smallint \limits_0^\pi - \cos nx} \right]\) 

\( = \frac{2}{\pi }\left[ {\frac{{ - \pi \;}}{n}\cos n\pi + \frac{1}{{{n^2}}}\left[ {\sin nx} \right]_0^\pi } \right]\) 

\( = \frac{2}{\pi }\left[ {\frac{{ - \pi }}{n}{{\left( { - 1} \right)}^n} + \frac{1}{{{n^2}}}\left( 0 \right)} \right]\) 

\( = \frac{{ - 2}}{n}{\left( { - 1} \right)^n}\) 

\(f\left( x \right) = x = \mathop \sum \limits_{k = 1}^\infty \frac{{ - 2}}{n}{\left( { - 1} \right)^n}\sin kx\) 

\({a_2} = \frac{{ - 2}}{2}{\left( { - 1} \right)^2} = - 1\)  

\(u = \log \frac{{{x^2}}}{y}\)

\(\frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2}/y}}.\frac{{2x}}{y} = \frac{2}{x} \Rightarrow x\frac{{\partial u}}{{\partial x}} = 2\)

\(\frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2}/y}}.\frac{{ - {x^2}}}{{{y^2}}} = \frac{{ - 1}}{y} \Rightarrow y\frac{{\partial u}}{{\partial x}} = - 1\)

Now,

For any density function,

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

\( \Rightarrow \mathop \smallint \limits_0^3 kxdx = 1\)

\( \Rightarrow k = \frac{2}{9}\)

P(|X − 1.8| < 0.6) = P(1.2 < X < 2.4)

\( = \mathop \smallint \limits_{1.2}^{2.4} \frac{2}{9}xdx = \frac{1}{9}\left( {{{2.4}^2} - {{1.2}^2}} \right) = 0.48\)

Let X be the number of defectives when n items are packed into a box.

P(X = 0) = (0.99)ⁿ

P(X ≥ 1) = 1 – P(X = 0) = 1 − (0.99)ⁿ

P(X ≥ 1) < 0.5

⇒ 1 − (0.99)ⁿ < 0.5

⇒ n < log 0.5/ log 0.99 = 68.97

Given that,

\(\phi \left( {x,y} \right) = log\left( {{x^2} + {y^2}} \right)\)

\(f\left( z \right) = \phi \left( {x,y} \right) + i\psi \left( {x\;y} \right)\)

Where \(\phi \left( {x,y} \right)\) is potential function

\(\psi \left( {x,y} \right)\) is complex potential function

\(f'\left( z \right) = {\phi _x} + i{\psi _x}\)

By CR equations, \({\phi _y} = - {\psi _x}\)

\(\phi \left( {x,\;y} \right) = log\;\left( {{x^2} + {y^2}} \right)\)

\({\phi _x} = \frac{1}{{\left( {{x^2} + {y^2}} \right)}}\left( {2x} \right),\;{\phi _y} = \frac{{2y}}{{\left( {{x^2} + {y^2}} \right)}}\)

\( \Rightarrow f'\left( z \right) = \left( {\frac{{2x}}{{{x^2} + {y^2}}}} \right) - i\left( {\frac{{2y}}{{{x^2} + {y^2}}}} \right)\)

By Milne-Thomson’s method, we express \(\frac{{df}}{{dz}}\) in terms of z, on replacing x by z and y by 0.

\( \Rightarrow f'\left( z \right) = \frac{{2z}}{{{z^2}}} - (i\times 0)\)

\( \Rightarrow f'(z) = \frac{2}{z}\)

\( \Rightarrow f\left( z \right) = \smallint \frac{2}{z}dz + c\)

\( \Rightarrow f\left( z \right) = 2\log z + c\)

Given differential equation is,

t² y’ – y² – yt = 0

Dividing the above equation by t² gives,

\(y' = \frac{{{y^2}}}{{{t^2}}} + \frac{y}{t}\)

\( \Rightarrow y' = {\left( {\frac{y}{t}} \right)^2} + \frac{y}{t}\)

Put,

⇒ y = tz

⇒ y’ = z + yz’

Now the differential equation becomes

z + tz’ = z² + z

\( \Rightarrow t\frac{{dz}}{{dt}} = {z^2}\)

\( \Rightarrow \frac{{dz}}{{{z^2}}} = \frac{{dt}}{t}\)

\( \Rightarrow \smallint \frac{{dz}}{{{z^2}}} = \smallint \frac{{dt}}{t} + C\)

\( \Rightarrow \frac{{ - 1}}{2} = \ln t + C\)

\(z = \frac{{ - 1}}{{\ln t + C}}\)

\( \Rightarrow y = \frac{{ - t}}{{\ln t + C}}\)

y’’ + 3y’ = e^2t

AE: D² + 3D = 0

⇒ (D + 3) D = 0

⇒ D = 0, -3

CF is y = Ae^-3t + B

\(PI = \frac{1}{{D\left( {D + 3} \right)}}{e^{2t}} = \frac{1}{{2\left( 5 \right)}}{e^{2t}} = \frac{{{e^{2t}}}}{{10}}\)

The solution is,

\(y\left( t \right) = A{e^{ - 3t}} + B + \frac{{{e^{2t}}}}{{10}}\)

y(0) = 1

\( \Rightarrow 1 = A + B + \frac{1}{{10}} \Rightarrow A + B = \frac{9}{{10}}\)

\(y'\left( t \right) = - 3A{e^{ - 3t}} + \frac{2}{{10}}{e^{2t}}\)

y'(0) = 0

\( \Rightarrow 0 = - 3A + \frac{2}{{10}}\)

\(\Rightarrow A = \frac{1}{{15}}\)

\( \Rightarrow B = \frac{9}{{10}} - \frac{1}{{15}} = \frac{5}{6}\)

\(y\left( t \right) = \frac{1}{{15}}{e^{ - 3t}} + \frac{1}{{10}}{e^{2t}} + \frac{5}{6}\)

At t = 1,

\(y\left( 1 \right) = \frac{1}{{15}}{e^{ - 3}} + \frac{1}{{10}}{e^2} + \frac{5}{6}\)

⇒ y(1) = 1.57

\(I = \mathop \smallint \limits_C^\; \frac{{f\left( z \right)}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}dz\)

\(f\left( z \right) = \sin \frac{{\pi z}}{2} + \cos \frac{{\pi z}}{2}\)

According to Cauchy’s residue theorem,

\(I = \mathop \smallint \limits_C^\; \frac{{f\left( z \right)}}{{\left( {z - 1} \right)\left( {z - 2} \right)}} = 2\pi i\;\left[ {sum\;of\;residues} \right]\)

Poles are, z = 1 and z = 2.

Residue at z = 1,

\(\mathop {{\rm{lt}}}\limits_{z \to 1} \frac{{f\left( z \right)}}{{\left( {z - 2} \right)}} = \mathop {{\rm{lt}}}\limits_{z \to 1} \frac{{\sin \frac{{\pi z}}{2} + \cos \frac{{\pi z}}{2}}}{{\left( {z - 2} \right)}}\)

= -1

Residue at z = 2

\(\mathop {{\rm{lt}}}\limits_{z \to 2} \frac{{f\left( z \right)}}{{\left( {z - 1} \right)}} = \mathop {{\rm{lt}}}\limits_{z \to 2} \frac{{\sin \frac{{\pi z}}{2} + \cos \frac{{\pi z}}{2}}}{{\left( {z - 1} \right)}}\)

= -1

The expected value of X is,

\(E\left( X \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( x \right)dx = \mathop \smallint \limits_0^1 2{x^2}dx = \frac{2}{3}\)

The expected value of X² is,

\(E\left( {{X^2}} \right) = \mathop \smallint \limits_{ - \infty }^\infty {x^2}f\left( x \right)dx = \mathop \smallint \limits_0^1 2{x^3}dx = \frac{1}{2}\)

The variance of X is,

\(Var\left( X \right) = E\left( X \right) - {\left( {E\left( X \right)} \right)^2} = \frac{1}{2} - {\left( {\frac{2}{3}} \right)^2} = \frac{1}{{18}}\)

Y = −2X + 3

The expectation of Y is,

\(E\left( Y \right) = - 2E\left( X \right) + 3 = - 2 \times \frac{2}{3} + 3 = \frac{5}{3}\)

\(Var\left( Y \right) = {\left( { - 2} \right)^2}Var\left( X \right) = 4 \times \frac{1}{{18}} = \frac{2}{9}\)

P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.7

\(P\left( {A{\rm{|}}B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{1}{3}\)

\(P\left( {B{\rm{|}}A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \frac{1}{5}\)

\(P\left( {A{\rm{|}}A \cup B} \right) = \frac{{P\left( A \right)}}{{{\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right)}} = \frac{5}{7}\)

P(A|A ∩ B) = P(A ∩ B)/P(A ∩ B) = 1

\(P\left( {A \cap B{\rm{|}}A \cup B} \right) = \frac{{P\left( {A \cap B} \right)}}{{{\rm{P}}({\rm{A}} \cup {\rm{B}}}} = \frac{1}{7}\)

V₁, V₂ are orthogonal vectors of unit length as shown below.

V₁^t V₂ = -cos x cos y sin y + cos x cos y sin y = 0

||V₁||² = cos² x ⋅ cos² y + cos² x sin² y + sin² x

= cos² x + sin² x = 1

||V₂||² = sin² y + cos² y = 1

The third column is perpendicular to each of the first two columns, so it is parallel to

\({V_3} = {V_1} \times {V_2} = \left[ {\begin{array}{*{20}{c}}{\cos x\cos y}\\{\sin x}\\{\cos x\sin y}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{ - \sin y}\\0\\{\cos y}\end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}}{\sin x\cos y}\\{ - \cos x}\\{\sin x\sin y}\end{array}} \right]\)

This is actually a unit vector itself, so the third column could be either V₃ or –V₃

So, the third column will be

\(\left[ {\begin{array}{*{20}{c}}{\sin x\cos y}\\{ - \cos x}\\{\sin x\sin y}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - \sin x\cos y}\\{\cos x}\\{ - \sin x\sin y}\end{array}} \right]\)

\(A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 1}&2&{ - 1}\\0&{ - 1}&1\end{array}} \right]\)

\(\left[ {A - \lambda I} \right] = \left[ {\begin{array}{*{20}{c}}{1 - \lambda }&{ - 1}&0\\{ - 1}&{2 - \lambda }&{ - 1}\\0&{ - 1}&{1 - \lambda }\end{array}} \right]\)

|A - λI|= 0

\( \Rightarrow \left| {\begin{array}{*{20}{c}}{1 - \lambda }&{ - 1}&0\\{ - 1}&{2 - \lambda }&{ - 1}\\0&{ - 1}&{1 - \lambda }\end{array}} \right| = 0\)

⇒ (1 - λ) (2 - λ) (1 - λ) – 2(1 - λ) = 0

⇒ (1 - λ) (- λ) (3 - λ) = 0

⇒ λ = 0, 1, 3

Eigen values of A are = 0, 1, 3

For λ₁ = 0:

A x₁ = λ₁x₁

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 1}&2&{ - 1}\\0&{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = 0\)

⇒ x₁ – x₂ = 0, -x₁ + 2x₂ – x₃ = 0, -x₂ + x₃ = 0

⇒ x₁ = x₂ = x₃

Eigen vector \(= \left[ {\begin{array}{*{20}{c}}K\\K\\K\end{array}} \right]\)

For K = 1, \({x_1} = \left[ {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right]\)

For λ₂ = 1

Ax₂ = λ₂x₂

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 1}&2&{ - 1}\\0&{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\)

⇒ x₁ – x₂ = x₁, -x₁ + 2x₂ – x₃ = x₂, -x₂ + x₃ = x₃

⇒ -x₁ = x₃ and x₂ = 0

Eigen vector \(= \left[ {\begin{array}{*{20}{c}}K\\0\\{ - K}\end{array}} \right]\)

For K = 1, \({x_2} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right]\)

For λ₃ = 3

Ax₃ = λ₃x₃

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 1}&2&{ - 1}\\0&{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3{x_1}}\\{3{x_2}}\\{3{x_3}}\end{array}} \right]\)

⇒ x₁ – x₂ = 3x₁, -x₁ + 2x₂ – x₃ = 3x₂, -x₂ + x₃ = 3x₃

⇒ x₂ = -2x₃, x₁ = x₃

Eigen vector \( = \left[ {\begin{array}{*{20}{c}}K\\{ - 2K}\\K\end{array}} \right]\)

For K = 1, \({x_3} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right]\)

The Eigen vectors of (A³ + 5I) are same as the Eigen vectors of A.

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = b₁

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = b₂

…

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = b_m

The above equations containing the n unknowns x₁, x₂, …, x_n. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & \ldots & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & \ldots & {{a}_{2n}} \\ \ldots & \ldots & \ldots & \ldots \\ {{a}_{m1}} & {{a}_{m2}} & \ldots & {{a}_{mn}} \\ \end{matrix} \right]~and~\left[ A\text{ }\!\!|\!\!\text{ }B \right]=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & \ldots & {{a}_{1n}} & {{b}_{1}} \\ {{a}_{21}} & {{a}_{22}} & \ldots & {{a}_{2n}} & {{b}_{2}} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ {{a}_{m1}} & {{a}_{m2}} & \ldots & {{a}_{mn}} & {{b}_{m}} \\ \end{matrix} \right]\)

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of augmented matrix and it is equal to number of unknowns, then system is consistent and there is a unique solution.

Rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of augmented matrix and it is less than the number of unknowns, then system is consistent and there are infinite number of solutions.

Rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to rank of augmented matrix, then system is inconsistent, and it has no solution.

Rank of A ≠ Rank of augmented matrix

Calculation:

Case a: x₁ + x₂ = 1, 2x₁ – x₂ = 0

The given system of equations can be represented in a matrix form as shown below.

\(A=\left[ \begin{matrix} 1 & 1 \\ 2 & -1 \\ \end{matrix} \right],~X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ \end{matrix} \right],~B=\left[ \begin{matrix} 1 \\ 0 \\ \end{matrix} \right]\)

The Augmented matrix can be written by

\(\left[ A\text{ }\!\!|\!\!\text{ }B \right]=\left[ \begin{matrix} 1 & 1 \\ 2 & -1 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ 0 \\ \end{matrix} \right]\)

R₂ → R₂ – 2R₁

\(=\left[ \begin{matrix} 1 & 1 \\ 0 & -3 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ -2 \\ \end{matrix} \right]\)

Rank of A = Rank of augmented matrix = n

So, it has unique solution.

Case b: x₁ + x₂ = 1, 2x₁ + 2x₂ = 2

The given system of equations can be represented in a matrix form as shown below.

\(A=\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \\ \end{matrix} \right],~X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ \end{matrix} \right],~B=\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]\)

The Augmented matrix can be written by

\(\left[ A\text{ }\!\!|\!\!\text{ }B \right]=\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ 2 \\ \end{matrix} \right]\)

R₂ → R₂ – 2R₁

\(=\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ 0 \\ \end{matrix} \right]\)

Rank of A = Rank of augmented matrix < n

So, it has infinitely many solutions.

Case c: x₁ + x₂ = 1, x₁ + x₂ = 0

The given system of equations can be represented in a matrix form as shown below.

\(A=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right],~X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ \end{matrix} \right],~B=\left[ \begin{matrix} 1 \\ 0 \\ \end{matrix} \right]\)

The Augmented matrix can be written by

\(\left[ A\text{ }\!\!|\!\!\text{ }B \right]=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ 0 \\ \end{matrix} \right]\)

R₂ → R₂ – R₁

\(=\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \\ \end{matrix}\text{ }\!\!|\!\!\text{ }\begin{matrix} 1 \\ -1 \\ \end{matrix} \right]\)

Rank of A ≠ Rank of augmented matrix

\(\mathop \smallint \limits_0^{2a} f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}}{2\mathop \smallint \limits_0^a f\left( x \right)dx,\;\;f\left( {2a - x} \right) = f\left( x \right)}\\{0,\;\;f\left( {2a - x} \right) = - f\left( x \right)}\end{array}} \right.\)

From the property of integration as mentioned above, the given integral can be reduced to

\( = 2\mathop \smallint \limits_0^\pi \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}{{\sin }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9{{\sec }^2}\theta + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9\left( {1 + {{\tan }^2}\theta } \right) + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9 + 10{{\tan }^2}\theta }}} \right)d\theta \)

Put tan θ = t

⇒ sec²θ dθ = dt

\( = 12\mathop \smallint \limits_0^\infty \left( {\frac{1}{{9 + 10{t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}}\mathop \smallint \limits_0^\infty \left( {\frac{1}{{\frac{9}{{10}} + {t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}} \times \frac{1}{{\frac{3}{{\sqrt {10} }}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{t}{{\frac{3}{{\sqrt {10} }}}}} \right)} \right]_0^\infty \)

\( = \frac{4}{{\sqrt {10} }}\left[ {\frac{\pi }{2} - 0} \right] = \frac{{2\pi }}{{\sqrt {10} }}\)

\(\frac{{{\partial ^2}f}}{{\partial x\partial y}}\) at (0, 0) can be calculated as

\({f_{xy}}\left( {0,\;0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{f_y}\left( {h,\;0} \right) - {f_y}\left( {0,\;0} \right)}}{h}\)

\({f_y}\left( {0,\;0} \right) = \mathop {\lim }\limits_{k \to 0} \frac{{f\left( {0,\;k} \right) - f\left( {0,\;0} \right)}}{k} = 0\)

\({f_y}\left( {h,\;0} \right) = \mathop {\lim }\limits_{k \to 0} \frac{{f\left( {h,\;k} \right) - f\left( {h,\;0} \right)}}{k}\)

\(= \mathop {\lim }\limits_{k \to 0} \frac{{hk\left( {{h^2} - {k^2}} \right)}}{{k\left( {{h^2} + {k^2}} \right)}} = h\)

\({f_{xy}}\left( {0,\;0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{h - 0}}{h} = 1\)

\(\frac{{{\partial ^2}f}}{{\partial y\partial x}}\) at (0, 0) can be calculated as

\({f_{yx}}\left( {0,\;0} \right) = \mathop {\lim }\limits_{k \to 0} \frac{{{f_x}\left( {0,\;k} \right) - {f_x}\left( {0,\;0} \right)}}{k}\)

\({f_x}\left( {0,\;0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h,\;0} \right) - f\left( {0,\;0} \right)}}{h} = 0\)

\({f_x}\left( {0,\;k} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h,\;k} \right) - f\left( {0,\;k} \right)}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{hk\left( {{h^2} - {k^2}} \right)}}{{k\left( {{h^2} + {k^2}} \right)}} = - k\)

y⁽⁴⁾ = -sin t + cos t

\(\smallint {y^{\left( 4 \right)}}dy = \smallint \left( { - \sin t + \cos t} \right)dt + c\)

⇒ y’’’(t) = cos t + sin t + c

y’’’(0) = 7 = 1 + 0 + c ⇒ c = 6

y’’’(t) = cos t + sin t + 6

\(\smallint y'''\left( t \right)dy = \smallint \left( {\cos t + \sin t + 6} \right)dt + c\)

⇒ y’’(t) = sin t – cos t + 6t + c

y’’(0) = 0 – 1 + c = -1 ⇒ c = 0

⇒ y’’(t) = sin t – cos t + 6t

\(\smallint y''\left( t \right)dy = \smallint \left( {\sin t - \cos t + 6t} \right)dt + c\)

⇒ y’(t) = -cos t – sin t + 3t² + c

y’(0) = -1 = -1 – 0 + c ⇒ c = 0

⇒ y’(t) = -cos t – sin t + 3t²

\(\smallint y'\left( t \right)dy = \smallint \left( { - \cos t - \sin t + 3{t^2}} \right) + c\)

⇒ y(t) = -sin t + cos t + t³ + c

y(0) = 0 = 1 + c ⇒ c = -1

y(t) = -sin t + cos t + t³ – 1

\(y\left( {\frac{\pi }{2}} \right) = - 1 + 0 + {\left( {\frac{\pi }{2}} \right)^3} - 1 = 1.876\)

t²y’’ – 3ty’ + 4y = 0

⇒ D(D - 1) y – 3Dy + 4y = 0

⇒ [D(D - 1) – 3D + 4] y = 0

⇒ [D² – 4D + 4] y = 0

D² – 4D + 4 = 0

⇒ D = 2, 2

Now, the solution is

y = [c₁ + c₂x] e^2x

y = [c₁ + c₂ ln t] e^{2 ln t}

⇒ y(t) = [c₁ + c₂ ln t] t²

y(1) = -2

⇒ -2 = c₁ (1) ⇒ c₁ = -2

y’(t) = [c₁ + c₂ ln t] (2t) + c₂(t)

y’(1) = 1

⇒ 1 = (-2)(2) + c₂(1) ⇒ c₂ = 5

Now, the solution is

y(t) = [-2 + 5 ln t] t²

Given equation is

\(\left( {2{D^2} + 5DD' + 2{{D'}^2}} \right)z = 0\)

Where \(D = \frac{{\partial z}}{{\partial x}},\;D' = \frac{{\partial z}}{{\partial y}}\)

Let D/D¹ = m

⇒ 2 m² + 5m + 2 = 0

⇒ m = -2, -1/2

The complete solution is

\(\begin{array}{l} Z = {f_1}\left( {y - 2x} \right) + {f_2}\left( {y - \frac{1}{2}x} \right)\\ \Rightarrow z = {f_1}\left( {y - 2x} \right) + {f_2}\left( {2y - x} \right) \end{array}\)