Engineering Mathematics Test 2

\({\rm{Let\ A}} = \left[ {\begin{array}{*{20}{c}}4&2\\1&3\end{array}} \right]\)

Characteristic equation of A is |A - λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}}{4 - {\rm{\lambda }}}&2\\1&{3 - {\rm{\lambda }}}\end{array}} \right| = 0\) 

⇒ λ² – 7λ + 10 = 0 ⇒ λ = 2, 5

\(Y\left( s \right) = L\left( {y\left( t \right)} \right) = \frac{5}{{2\left( {s - 1} \right)}} - \frac{2}{{s - 2}} + \frac{1}{{2\left( {s - 3} \right)}}\)

Apply inverse Laplace transform,

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

Differentiate with respect to ‘t’.

\( \Rightarrow y'\left( t \right) = \frac{5}{2}{e^t} - 4{e^{2t}} + \frac{3}{2}{e^{3t}}\)

\(\begin{array}{l} y\left( 0 \right) = \frac{5}{2} - 2 + \frac{1}{2} = 1\\ y'\left( 0 \right) = \frac{5}{2} - 4 + \frac{3}{2} = 0 \end{array}\)

⇒ y(0) + y'(0) = 1

The order of differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative accruing in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

\({\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^{\frac{1}{2}}} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^{\frac{1}{3}}}\)

Highest derivative in the given differential equation is 4

Hence order is 4.

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\({\left[ {{{\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)}^{\frac{1}{2}}}} \right]^6} = {\left[ {{{\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]}^{\frac{1}{3}}}} \right]^6}\)

\(\Rightarrow {\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^3} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^2}\)

Now the differential equation is free from radicals

Degree of highest derivative = 3

Order and degree of the given differential equation = 4 and 3

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{ - \sin {\rm{x}}}}{{2\sin {\rm{x}} + {\rm{x}}\cos {\rm{x}}}}} \right){\rm{}}\left( {\frac{0}{0}{\rm{form}}} \right)\)

Applying L – Hospital Rule, we get

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{ - \cos {\rm{x}}}}{{2\cos {\rm{x}} + \cos {\rm{x}} - {\rm{x}}\sin {\rm{x}}}}} \right)\)

= -1/3

Concept: 

A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

Calculation:

For the function to be continuous,

\(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = f\left( 3 \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)\)

\(\mathop {\lim }\limits_{x \to {3^ - }} \left( {-4 + x^2} \right) = \mathop {\lim }\limits_{x \to 3} \left( {-4 +x^2} \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {-2x +a} \right)\)

⇒ - 4 + 3² = - 2 × 3 + a ⇒ 5 = - 6 + a

⇒ a = 11

Concept:

The divergence of any vector field  \(\vec A\) is defined as:

\(Div= \vec \nabla .\vec A\)

The nabla operator is defined as:

\(\vec \nabla ={\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}}}+\hat k\frac{\partial }{\partial z}\)

Calculation:

\(f = {x^3}y\vec i - \left( {{z^2} - 2y} \right)\vec j + 5{y^2}z\vec k\)

\(Div\left( f \right) = \frac{\partial }{{\partial x}}\left( {{x^3}y} \right) + \frac{\partial }{{\partial y}}\left( { - \left( {{z^2} - 2y} \right)} \right) + \frac{\partial }{{\partial z}}\left( {5{y^2}z} \right)\)

= 3x²y + (+2) + (5y²)

= 3x²y + 5y² + 2

At (2, 3, 4)

Div (f) = 3(2)²(3) + 5(3)² + 2

∴ Div (f) = 36 + 45 + 2 = 83

f(x) = 3x3 – 40.5x2 + 180x + 7

f’(x) = 9x2 – 81x + 180 = 0

put f’(x) = 0 to find the point at which maximum and minimum value exists

9x2 – 81x + 180 = 0

x2 – 9x + 20 = 0

(x – 5)(x – 4 ) = 0

∴ x = 5 or x =4

Interval [4, 5]

f'’(x) = 2x – 9

put x = 4

f’’(x) = -1 < 0 (maximum value might exist)

put x = 5

f’’(x) = 1 > 0 (minimum value might exist)

Also check border values:

The minimum value is 269.5

Concept:

\(S.D = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \)

Calculation:

\(E\left( X \right) = 2\left( {\frac{1}{3}} \right) + \left( { - 1} \right)\left( {\frac{2}{3}} \right)\)

∴ E(X) = 0

Now,

\(E\left( {{X^2}} \right) = \sum x_i^2P\left( {{x_i}} \right)\)

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

∴ E(X²) = 2

We know that,

\(S.D\;\left( \sigma \right) = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \) 

\(\therefore S.D\;\left( \sigma \right) = \sqrt {{2^2} - 2} \)

∴ σ = √2 = 1.414

Concept:

\(\begin{array}{l} \int\limits_0^3 {f(x)dx} = \frac{h}{3}\left[ {\left( {\mathop y\nolimits_0 + \mathop y\nolimits_n } \right) + 4(\mathop y\nolimits_1 + \mathop y\nolimits_3 + ......) + 2(\mathop y\nolimits_2 + \mathop y\nolimits_4 + ......)} \right]\\ \end{array}\)

Calculation:

\(\begin{array}{l} \int\limits_0^3 {f(x)dx} = \frac{h}{3}\left[ {\left( {\mathop y\nolimits_0 + \mathop y\nolimits_3 } \right) + 4(\mathop y\nolimits_1 ) + 2(\mathop y\nolimits_2 )} \right] = \frac{1}{3}\left[ {\left( {44 + 2} \right) + 4(5) + 2(16)} \right] = 32.67\\ \end{array}\)

For a probability density function,

\(\begin{array}{l}\mathop \smallint \limits_{ - \infty }^\infty p\left( x \right)dx = 1\\\Rightarrow \mathop \smallint \limits_{ - \infty }^\infty k\;{e^{ - \left| x \right|}}dx = 1\end{array}\)

\(\begin{array}{l}\Rightarrow \mathop \smallint \limits_{ - \infty }^0 k{e^x}\;dx + \mathop \smallint \limits_0^\infty k{e^{ - x}}\;dx = 1\\\Rightarrow \left[ {k{e^x}} \right]_{ - \infty }^0 + \left[ { - k{e^{ - x}}} \right]_0^\infty = 1\end{array}\)

⇒ k (1) + [-k (0 – 1)] = 1

⇒ k + k = 1

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 2\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

If f(x)  is odd function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

Calculation:

Let I = \(\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \rm \dfrac{sin^3 \ x \ e^{-x^2}}{x^4}dx\)

Let f(x) = \( \dfrac{sin^3 \ x \ e^{-x^2}}{x^4}\)

Replaced x by -x, 

⇒ f(-x) = \(\rm \dfrac{sin^3 \ (-x) \ e^{-{(-x)^2}}}{(-x)^4}\)

As we know sin (-θ) = - sin θ

= \( \dfrac{-sin^3 \ x \ e^{-x^2}}{x^4}\)

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

Length of curve is given by

\(L = \mathop \smallint \limits_{x = a}^{x = b} \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx\)

\(y = \frac{2}{3}{x^{\frac{3}{2}}}\)

\(\frac{{dy}}{{dx}} = \sqrt x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)^2} = x\) ---(1)

\(L = \mathop \smallint \limits_{x = 0}^{x = 3} \sqrt {1 + x} \;dx\)

\(\left[ {\frac{{{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3\)

\(= \frac{2}{3}\left[ {{{\left( {x + 1} \right)}^{\frac{3}{2}}}} \right]_0^3\)

\(= \frac{2}{3}\left[ {{4^{\frac{3}{2}}} - 1} \right]\)

Rolle’s Theorem:

Suppose f(x) is continuous on [a, b] differentiable on (a, b) and f(a) = f(b). Then there exists some point c ϵ [a, b] such that f'(c) = 0.

Calculation:

The given function is continuous and differentiable at x = 0

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ - }} \end{array}f\left( x \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ + }} \end{array}f\left( x \right)\)

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}f\left( {1 + x} \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}\left( {1 - x} \right)\left( {px + q} \right)\)

⇒ 1 = q

The function to be differentiable at x = 0

f'(x = 0) = f'(x = 0⁺)

⇒ 1 = (-1) (px + q) + (1 – x) p

At, x = 0

⇒ 1 = -q + p ⇒ p = 1 + q = 2

Explanation:

\(\nabla \cdot \left( {\phi \vec A} \right) = \left( {\nabla \phi } \right) \cdot \vec A + \phi \left( {\nabla \cdot \vec A} \right)\)

\(\nabla \cdot \left( {\frac{1}{{{r^2}}} \cdot \vec r} \right) = \left( {\nabla \frac{1}{{{r^2}}}} \right) \cdot \vec r + \frac{1}{{{r^2}}}\left( {\nabla \cdot \vec r} \right)\)

\(= \left[ {\frac{{ - 2\vec r}}{{{r^4}}}} \right] \cdot \vec r + \frac{3}{{{r^2}}}\;\;\;\;\;\;\;\;\;\;\;\left[ {\because \nabla \;\left( {f\left( r \right)} \right) = \frac{{f'\left( r \right)}}{r}\vec r} \right]\)

\( = \frac{{ - 2}}{{{r^2}}} + \frac{3}{{{r^2}}}\;\;\;\;\;\;\;\;\;\;\left\{ {\vec r \cdot \vec r = {r^2}} \right\}\)

\(= \frac{1}{{{r^2}}}\)

\({\nabla ^2}\left[ {\nabla \cdot \frac{{\vec r}}{{{r^2}}}} \right] = {\nabla ^2}\left[ {\frac{1}{{{r^2}}}} \right]\)

\( = \frac{6}{{{r^4}}} + \frac{2}{r}\left( {\frac{{ - 2}}{{{r^3}}}} \right)\)

\(\left[ {{\nabla ^2}\left\{ {f\left( r \right)} \right\} = r''\left( r \right) + \frac{2}{r}f'\left( r \right)} \right]\)

\(= \frac{6}{{{r^4}}} - \frac{4}{{{r^4}}}\)

\(= \frac{2}{{{r^4}}} = 2{r^{ - 4}}\)

Concept:

Write auxillary equation by replacing

\(\frac{{{d^n}y}}{{d{x^n}}} = {m^n}\)

Equate to 0 and solve for (m) (f(m) = 0)

→ for repeated roots m₁ = m₂ = m (real)

Complementary function (CF)

CF = (C₁ + C₂ x) e^mx

→ Particular integral (PI) when f(x) = e^ax

\(PI = \frac{{{e^{ax}}}}{{f\left( m \right)}}\)

Replace m by a

Solution ⇒ y = CF + PI

Calculation:

m² + 4m + 4 = 0

(m + 2)² = 0 ⇒ m = -2, -2

C F = (C₁ + C₂ x) e^-2x

\(PI = \frac{{{e^{2x}}}}{{{m^2} + 4m + 4}} = \frac{{{e^{2x}}}}{{16}}\)

∴ y = (C₁ + C₂ x) e^-2x + e^2x / 16

y(0) = 4 ⇒ 4 = C₁ + 1 / 16 ⇒ C₁ = 63 / 16

y = (C₁ + C₂ x) e^-2x + e^2x / 16

\(y'\left( x \right) = - 2\left( {{C_1} + {C_2}x} \right){e^{ - 2x}} + \frac{{2{e^{2x}}}}{{16}} + {C_2}{e^{ - 2x}}\)

\(y'\left( 0 \right) = - 2{C_1} + \frac{2}{{16}} + {C_2}\) 

\(9 = - 2 \times \frac{{63}}{{16}} + \frac{2}{{16}} + {C_2}\) 

\({C_2} = 9 + \frac{{63}}{8} - \frac{1}{8} = 9 + \frac{{62}}{8}\) 

C₂ = 16.75

C₁ = 63 / 16 = 3.9375

Now,

\(y = \left( {3.9375 + 1675x} \right){e^{ - 2x}} + \frac{{{e^{2x}}}}{{16}}\) 

\(\therefore y\left( 1 \right) = \frac{{\left( {3.9375 + 16.75} \right)}}{{{e^2}}} + \frac{{{e^2}}}{{16}}\)

∴ y(1) = 3.26 

Concept:

According to Newton-Raphson Method

\({X_{n + 1}} = {X_n} - \frac{{f\left( {{X_n}} \right)}}{{f'\left( {{X_n}} \right)}}\)

Calculation:

f(x) = e^x – 1

f’(x) = e^x

x₀ = 1

Iteration 1:

At x₀ = 1, f(x) = 1.718

f'(x) = 2.718

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

\( = 1 - \frac{{1.718}}{{2.718}} = 0.3678\)

Iteration 1:

At x₁ = 0.3678, f(x) = 0.4445

f'(x) = 1.4445

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

\( = 0.3678 - \frac{{0.4447}}{{1.4447}} = 0.06005\)

By inspection, the actual solution for the given equation is, x = 0

Therefore, the error = 0.06005

Concept:

\({\rm{Number\;of\;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)

where, b is the upper limit, a is the lower limit, h is the step size

According to the trapezoidal rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)

And,

\(Absolute\;percentage\;error = \frac{{\left| {true\;value - approximate\;value} \right|}}{{true\;value}}\)

Calculation:

Given, Number of interval = 1

a = 0, \({\rm{b\;}} = {\rm{\;}}\frac{\pi }{2}\)

\(\begin{array}{l}{\rm{Number\;of\;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}\\ \Rightarrow 1 = \frac{{\frac{{\rm{\pi }}}{2} - 0}}{{\rm{h}}}{\rm{\;}}\\\therefore {\rm{h}} = \frac{{\rm{\pi }}}{2}\end{array}\)

True value

\(\mathop \smallint \nolimits_0^{\pi /2} \left( {8 + 4\cos x} \right)dx = \;\left[ {8x + 4sinx} \right]_0^{\frac{\pi }{2}}\)

\(\Rightarrow {\rm{True\;value}} = \left[ {\left( {8 \times \frac{\pi }{2}} \right) + \left( {4 \times \sin \left( {\frac{\pi }{2}} \right)} \right)} \right]\;\)

⇒ True value = 4π + 4 = 16.56

Approximate value

By trapezoidal rule

\(h = \frac{{\left( {\frac{\pi }{2} - 0} \right)}}{1} = \frac{\pi }{2}\)

According to the trapezoidal rule

\(\begin{array}{l}\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\\\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {8{\rm{\;}} + 4{\rm{cosx}}} \right){\rm{dx\;}} = \frac{{\rm{\pi }}}{4}\left( {12 + 8} \right)\end{array}\)

⇒  Approximate value = 5π = 15.70

Now,

\(\begin{array}{l}Absolute\;percentage\;error = \frac{{\left| {true\;value - approximate\;value} \right|}}{{true\;value}}\\Absolute\;percentage\;error = \frac{{\left| {16.56 - 15.70} \right|}}{{16.56}} \times 100\end{array}\)

∴ Absolute percentage error = 5.19 %

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

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

|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.

Probability of the component to function (p) = 0.8

Probability of the component do not function (q) = 0.2

Minimum probability of functioning the system = 0.97

⇒ 1 – probability that no component functions ≥ 0.97

⇒ P(x = 0) ≤ 0.03

\( \Rightarrow {}_\;^n{C_0}{\left( p \right)^0}{\left( q \right)^n} \le 0.03\)

\( \Rightarrow {}_\;^n{C_0}{\left( {0.8} \right)^0}{\left( {0.2} \right)^n} \le 0.03\)

⇒ (0.2)ⁿ ≤ 0.03

The minimum value n so that the above condition holds = 3.

Explanation:

Given: \({x^2}{y^{11}} + 6x{y^I} + 6y = x\) 

Now,

D(D – 1)y + 6Dy + 6y = e^t, where \(D = \frac{d}{{dt}}\) 

Put x = e^t, t = ln x

⇒ (D² + 5D + 6) y = e^t

Auxillary Equation is D² + 5D + 6 = 0

Solving the above Equation we get roots as -2, -3

Now,

\(CF = {c_1}{e^{ - 2t}} + {c_2}{e^{ - 3t}}\)

\(CF = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}}\)

\(\therefore {\rm{The\;complementary\;function\;is\;}}\left( {C.F.} \right) = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}}\)

Now,

\(P.I = \frac{1}{{\left( {{D^2} + 5D + 6} \right)}}{e^t}\)

\(P.I. = \frac{{{e^t}}}{{\left( {{1^2} + 5 \times 1 + 6} \right)}} = \frac{{{e^t}}}{{12}}\)

\(\therefore P.I = \frac{x}{{12}}\)

\(\therefore {\rm{The\;particular\;integral\;is\;}}\frac{x}{{12}}\)

Now,

∴ The general solution is equal to C.F. + P.I.

\(\therefore y = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}} + \frac{x}{{12}}\)

Concept:

Line Integral:

Any integral which is to be evaluated along a closed curve path 'C' is called line integral. The line integral of a vector function \(\vec F\)along a curve 'C' is given by- curve can be closed or open.

\(\int_{c}^{}\vec{F}.\vec{dr}=\int_{c}^{}(F_1dx\;+\;F_2dy\;+\;F_3dz)\)

Calculation:

Given:

\(\mathop \oint \nolimits_C \left( {y\hat i - 3x\hat j} \right) \cdot \overrightarrow {dr} \) and y = x2 – 4x + 4

∴ dy = 2x dx – 4 dx = (2x - 4)dx

\(\vec F=y\hat i\;-\;3x\hat j\;\;and\;\;\overrightarrow {dr} = dx\;\hat i + dy\;\hat j\)

\(\)\(\therefore\;\mathop \oint \nolimits_C \left( {y\hat i - 3x\hat j} \right) \cdot \left( {dx\;\hat i + dy\;\hat j} \right)\)

\( \therefore\;\mathop \oint \nolimits_C ydx - 3x\;dy\)

\( \therefore\;\mathop \oint \nolimits_C \left( {{x^2} - 4x + 4} \right)dx - 3x\left( {2x - 4} \right)dx\)

\(\therefore\;\mathop \oint \nolimits_C \left( { - 5{x^2} + 8x + 4} \right)dx\)

The point given is (0, 4) and (2, 0)

x varies from 0 → 2 and y varies 4 → 0.

\(\mathop \smallint \limits_{x = 0}^2 \left( { - 5{x^2} + 8x + 4} \right)dx\)

\( \therefore\;\left[ {\frac{{ - 5{x^3}}}{3} + 4{x^2} + 4x} \right]_0^2\)

\(\therefore\;\frac{{ - 5}}{3}\left( 8 \right) + 4\left( 4 \right) + 4\left( 2 \right)\)

\(\therefore\;\frac{{ - 40}}{3} + 24 \Rightarrow \frac{{32}}{3}\)

Concept:

The rank of a matrix is defined as the maximum number of linearly independent column vectors in the matrix or the maximum number of linearly independent row vectors in the matrix.

In terms of determinant, rank of a matrix can be viewed as m where m is the size of the largest non- zero m × m submatrix with non-zero determinant.

Calculation:  

\(P = \left[ {\begin{array}{*{20}{c}}1&1&{ - 1}\\2&{ - 3}&4\\3&{ - 2}&3\end{array}} \right]\;and\;Q = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}&{ - 1}\\6&{12}&6\\5&{10}&5\end{array}} \right]\)

\(Let\;X = P + Q = \;\left[ {\begin{array}{*{20}{c}}0&{ - 1}&{ - 2}\\8&9&{10}\\8&8&8\end{array}} \right]\)

\(\left| X \right| = \left| {\begin{array}{*{20}{c}}0&{ - 1}&{ - 2}\\8&9&{10}\\8&8&8\end{array}} \right|\)

|X| = 0 (72 – 80) – (–1) (64 – 80) – 2(64 – 72) 

∴ |X| = 0 + (–16) + 16 = 0

Rank of 3 × 3 matrix is zero.

∴ Rank(X) ≠ 3

\(\left| {\begin{array}{*{20}{c}}0&{ - 1}\\8&9\end{array}} \right| = 8 \ne 0\)

Concept:

Fourier series of f(x)

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos \frac{{n\pi x}}{L} + \mathop \sum \limits_{n = 1}^\infty {n_n}\sin \frac{{n\pi x}}{l}\)

\({a_0} = \frac{1}{l}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx,\;{a_n} = \frac{1}{l}\mathop \smallint \limits_{ - l}^l f\left( x \right)\cos \frac{{n\pi x}}{L}\;dx,\;{b_n} = \frac{1}{l}\;\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

Calculation:

\({a_0} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)dx = \frac{1}{\pi }\;\mathop \smallint \limits_0^\pi \sin x\;dx = \frac{2}{\pi }\)

\({a_n} = \frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin x\cos nx\;dx = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi [\sin \left( {n + 1} \right)x - \sin \left( {n - 1} \right)x]\;dx\)

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

= 0, n is odd and \(\frac{{ - 2}}{{\pi \left( {{n^2} - 1} \right)}}\), n is even

Now,

\({b_n} = \frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin x\sin n\;x\;dx\)

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

\({b_n} = \frac{1}{{2\pi }}\left[ {\frac{{\sin \left( {n - 1} \right)x}}{{n - 1}} - \frac{{\sin \left( {n + 1} \right)x}}{{n + 1}}} \right]_0^\pi = 0\left( {n \ne 1} \right)\)

When n = 1

\({b_1} = \frac{1}{2}\)

\(\therefore f\left( x \right) = \frac{1}{\pi } - \frac{2}{\pi }\left[ {\frac{{\cos 2x}}{{{2^2} - 1}} + \frac{{\cos 4x}}{{{4^2} - 1}} + \ldots } \right] + \frac{1}{2}\sin x\)

\({\rm{Put\;}}x = \frac{\pi }{2}\)

\( \Rightarrow 1 = \frac{1}{\pi } - \frac{2}{\pi }\left( { - \frac{1}{{1.3}} + \frac{1}{{3.5}} - \frac{1}{{5.7}} + \ldots \infty } \right) + \frac{1}{2}\)

\(\therefore \frac{1}{{1.3}} - \frac{1}{{3.5}} + \frac{1}{{5.7}} - \ldots \infty = \frac{1}{4}\left( {\pi - 2} \right) = 0.285\)