Engineering Mathematics Test 8

3-D heat equation is given as below

\(\left( {\frac{{{\partial ^2}T}}{{d{x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}}} \right) + \frac{{Q\left( {x,t} \right)}}{K} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

For 1 – D & without heat generation:

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

Where α ÷ thermal diffusivity.

Wave equation is given by:

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right)\)      (2-D)

Laplace equation:

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\)     (3-D)

\({\nabla ^2}u = 0\)

​Poisson’s equation: 

\({\nabla ^2}V = - \frac{{{\rho _v}}}{\epsilon}\)

Concept:

Euler’s theorem for homogenous function

Case 1: If z = f(x,y) is homogenous function of degree ‘n’ in x and y then according to euler’s theorem,

• xzx + yzy = nz
• x2zxx + 2xyzxy + y2zyy = n(n – 1)z

Case 2: If z = f(x,y) is non-homogenous function however, u = g(z) is homogenous function of degree ‘n’ in x and y then according to euler’s theorem,

• \(xz_x\; + \;yz_y\; = \frac{{ng\left( z \right)}}{{g'\left( z \right)}} = E\left( z \right)\)
• x2zxx + 2xyzxy + y2zyy = E(z)[E’(z) – 1]

Calculation:

Given:

\(z = {\sin ^{ - 1}}\left( {\frac{{{x^3} + {y^3}}}{{x - y}}} \right)\)

As z is non-homogenous function, however, u = g(z) = sin z is homogenous function of degree 2 in x and y respectively.

\(xz_x\; + \;yz_y\; = \frac{{ng\left( z \right)}}{{g'\left( z \right)}} = E\left( z \right)\)

\(x{z_x} + \;y{z_y} = \frac{{2\sin z}}{{\cos z}} = 2\tan z\)

Concept:

If F(s) is the Laplace transform of f(t),

By using the first shifting rule

L [e^atf(t)] = F (s – a)

Application:

y(t) = te^-5t

L(t) = 1/s²

L [y(t)] = L (te^-5t)

\(L\left( {t{e^{ - 5t}}} \right) = \frac{1}{{{{\left( {s + 5} \right)}^2}}}\)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 7x\frac{{dy}}{{dx}} + 16y = 0\)

Put x = e^t

⇒ t = ln x

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} \Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)

Now, the above differential equation becomes

D (D – 1) y – 7 D y + 16 y = 0

⇒ D² y – D y – 7 D y + 16 y = 0

⇒ (D² – 8 D + 16) y = 0

Auxiliary equation:

(D² – 8 D + 16) = 0

⇒ D = 4

The solutions for the above roots of auxiliary equations are:

y(t) = (c₁ + c₂ t) e^4t

Explanation:

\(u = {\tan ^{ - 1}}\left( {\frac{{{y^3} - {x^3}}}{{{x^2} + {y^2} + xy}}} \right)\)

\(\tan u = \frac{{{y^3} - {x^3}}}{{{x^2} + {y^2} + xy}} = \frac{{\left( {y - x} \right)\left( {{x^2} + {y^2} + xy} \right)}}{{\left( {{x^2} + {y^2} + xy} \right)}}\)

Now,

\(\tan u = y\left[ {1 - \left( {\frac{x}{y}} \right)} \right]\) is a homogeneous function of degree ‘1’

\(\therefore x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \frac{{n\;f\left(u \right)}}{{f'\left( u \right)}} = 1 \times \frac{{\tan u}}{{{{\sec }^2}u}}\)

\(= \frac{{\tan u}}{{1 + {{\tan }^2}u}} = \frac{{\left( {y - x} \right)}}{{1 + {{\left( {y - x} \right)}^2}}}\)

\(= \frac{{\left( {3 - 1} \right)}}{{1 + {{\left( {3 - 1} \right)}^2}}} = \frac{2}{5} = 0.4\)

\(y'' + y = t\)

By applying Laplace transform on both sides,

\( \Rightarrow {s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + Y\left( s \right) = \frac{1}{{{s^2}}}\)

\(\Rightarrow {s^2}Y\left( s \right) - s + 2 + Y\left( s \right) = \frac{1}{{{s^2}}}\)

\(\Rightarrow Y\left( s \right)\left( {1 + {s^2}} \right) = \frac{1}{{{s^2}}} + s - 2\)

\(\Rightarrow Y\left( s \right) = \frac{1}{{{s^2}\left( {1 + {s^2}} \right)}} + \frac{s}{{\left( {1 + {s^2}} \right)}} - \frac{2}{{\left( {1 + {s^2}} \right)}}\)

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

\(\Rightarrow Y\left( s \right) = \frac{1}{{{s^2}}} - \frac{3}{{1 + {s^2}}} + \frac{s}{{1 + {s^2}}}\)

By applying inverse Laplace transform,

Given:

\(\frac{{dy}}{{dx}} + \frac{y}{x} = x\)

Comparing with

\(\frac{{dy}}{{dx}} + py = q\)

We get,

\(p = \frac{1}{x};q = x\)

\(TF = {e^{\smallint pdx}} = {e^{\smallint \frac{1}{x}dx}} = {e^{\ln x}} = x\)

Now,

General solution can be written as,

Y (IF) = ∫ q (IF) dxt C

Y (x) = ∫ x (x) dxt C

xy = ∫ x² dx + C

\(xy = \frac{{{x^3}}}{3} + C\)

At, x = 1 & y  =1

\(\left( 1 \right)\left( 1 \right) = \frac{{\left( 1 \right)}}{3} + C\)

\(\therefore C = \frac{2}{3}\)

\(\therefore xy = \frac{{{x^3}}}{3} + \frac{2}{3}\)

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

Explanation:

Let F(t) = e^-at – e^-bt

{L{F(t)} = L{e^-at – e^-bt}

\(= \frac{1}{{s + a}} - \frac{1}{{s - b}}\)

Now \(L\left\{ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right\} = \mathop \smallint \limits_s^\infty \left( {\frac{1}{{\left( {s + a} \right)}} - \frac{1}{{\left( {s + b} \right)}}} \right)ds\)

\( = \left[ {\log \left( {s - a} \right) - \log \left( {s + b} \right)} \right]_s^\infty \)

\(= \left[ {\log \left( {\frac{{s + a}}{{s + b}}} \right)} \right]_s^\infty \)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{s + a}}{{s + b}}} \right) - \log \left( {\frac{{s + a}}{{s + b}}} \right)\)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{1 + a/s}}{{1 + b/s}}} \right) - \log \left[ {\frac{{s + a}}{{s + b}}} \right]\)

\( = \log \left[ {\frac{{1 + 0}}{{1 + 0}}} \right] + \log {\left( {\frac{{s + a}}{{s + b}}} \right)^{ - 1}} = \log \left( {\frac{{s + b}}{{s + a}}} \right)\)

∵ By definition

\(L\left\{ {F\left( t \right)} \right\} = \mathop \smallint \limits_o^\infty {e^{ - st}}.F\left( t \right)dt\)

\(L\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right] = \mathop \smallint \limits_0^\infty {e^{ - st}}\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right]dt\)

Put s = 0

\(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt = \log \frac{b}{a}\)

Laplace transform of \(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt\) = Laplace transform of \(\log \frac{b}{a}\)

\( = \frac{1}{s}\log \frac{b}{a}\)

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:

Euler Cauchy Homogeneous linear equation:

\({x^n}\frac{{{d^n}y}}{{d{x^n}}} + {c_1}{x^{n - 1}}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + ... + {c_{n - 1}}x\frac{{dy}}{{dx}} + {c_n}y = f(x)\)

Take x = et ⇒ t = log x

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} \Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)

Calculation:

\({x^2}\frac{{{d^3}y}}{{d{x^3}}} - 4x\frac{{{d^2}y}}{{d{x^2}}} + 6\frac{{dy}}{{dx}} = 4\)

By multiplying with ‘x’ on both sides

\(\Rightarrow {x^3}\frac{{{d^3}y}}{{d{x^3}}} - 4{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 6x\frac{{dy}}{{dx}} = 4x\)

The given equations become,

⇒ [D (D – 1) (D – 2) – 4 D (D – 1) + 6D] y = 4x

⇒ (D³ – 3D² + 2D – 4D² + 4D + 6D) y = 4x

⇒ (D³ – 7D² + 12D) y = 4x

Auxiliary equation: D³ – 7D² + 12D = 0

⇒ D (D² – 7D + 12) = 0

⇒ D = 0, 3, 4

Complementary solution (CF) = C₁ + C₂ e^3t + C₃ e^4t

CF = C1 + C2 x³ + C₃x⁴