Engineering Mathematics Test 7

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Engineering Mathematics Test 7

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Question 1
1 / -0

If \(\frac{{{d^2}y}}{{d{t^2}}} + y = 0\) under the conditions y = 1, \(\frac{{dy}}{{dt}} = 0\), when t = 0, then y(π/2) is equal to

A
0-0

B
04-01

C
07-06

D
09-02

Solution

(D² + 1) y = 0
A.E. is (D² + 1) = 0
⇒ D = ± i
Solution is, y = (C₁ cos t + C₂ sin t)
Given that, y(0) = 1
⇒ 1 = (C₁ + 0) ⇒ C₁ = 1
y’ = -C₁ sin t + C₂ cos t
y’(0) = 0
⇒ 0 = 0 + C₂ ⇒ C₂ = 0
⇒ y = cos t
\(\Rightarrow y\left( {\frac{\pi }{2}} \right) = \cos \left( {\frac{\pi }{2}} \right) = 0\)
Question 2
1 / -0

Which of the following equations represents a one-dimensional wave equation?

A
\(\frac{{\partial u}}{{\partial t}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

B
\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

C
\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{\partial u}}{{\partial x}}\)

D
\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} + \frac{{{\partial ^2}u}}{{\partial {x^2}}} = 0\)

Solution

One dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)
Two-dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}\)
Important:
Heat equation: \(\frac{{\partial u}}{{\partial t}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)
Laplace equation: \(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\)
Question 3
1 / -0

Find the values of y(1) by solving the differential equation \(y\frac{{dy}}{{dx}} = 6{x^2} + 5,\;y\left( 0 \right) = 2\)

A
\( \pm \sqrt 2 \)

B
0

C
\( \pm \;3\sqrt 2 \)

D
± 3

Solution

\(y\frac{{dy}}{{dx}} = 6{x^2} + 5\)
⇒ y dy = (6x² + 5) dx
By integrating both the sides, we get
\( \Rightarrow \frac{{{y^2}}}{2} = 2{x^3} + 5x + C\)
⇒ y² = 4x³ + 10x + C
y(0) = 2
⇒ C = 4
Now, y² = 4x³ + 10x + 4
At x = 1, y² = 4 + 10 + 4 = 18
\( \Rightarrow y\left( 1 \right) = \sqrt {18} = \pm 3\sqrt 2 \)
Question 4
1 / -0

The solution of the partial differential equation \(\frac{{{\partial ^2}z}}{{\partial {y^2}}} + z = 0\) is when \(y = 0,z = {e^x}\;and\frac{{\partial z}}{{\partial y}} = {e^{ - x}}\)

A
z = e^-x sin y + e^x cos y

B
z = e^-x sin 2y + e^x cos 2y

C
z = e^x sin y + e^-x cos y

D
z = e^x sin 2y + e^-x cos 2y

Solution

Explanation:
\(\frac{{{\partial ^2}}}{{\partial {y^2}}} + z = 0\)
If z were function of y alone, the solution would have been z = A sin y + B cos y, where A and B are constants.
Since z is a function of x and y, A and B can be arbitrary functions of x.
Hence the solution of the given equations is,
z = f(x) sin y + ϕ(x) cos y
\(\frac{{\partial z}}{{\partial y}} = f\left( x \right)\cos y - \phi \left( x \right)\sin y\)
At, y = 0, z = e^x
⇒ e^x = ϕ(x)
At, \(y = 0,\frac{{\partial z}}{{\partial y}} = {e^{ - x}}\)
⇒ e^-x = f(x)
The solution is,
z = e^-x sin y + e^x cos y
Question 5
1 / -0

A third order ordinary differential equation with x, x ln x and x ln²x as linearly independent solutions is given by

A
x³y''' + 2x²y'' - xy' + y = 0

B
x³y''' + xy' - y = 0

C
x³y''' + 3x²y'' - 2xy' + 2y = 0

D
x³y''' + 4x²y'' + xy' - y = 0

Solution

Linear differential equations with variable coefficients can be reduced to linear differential equations with constant coefficients by suitable substitutions.
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
\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} \Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)
Given that, x, x ln x. and ln²x as linearly independent solutions.
y = C₁x + C₂x ln x + C₃x ln² x
put, t = ln x
⇒ x = e^t
⇒ y = C₁e^t + C₂t e^t + C₃t²e^t
⇒ y = (C₁ + C₂t + C₃t²) e^t
From the above equation, roots of auxiliary equation are, D = 1, 1, 1
Now, the auxiliary equation is,
(D - 1) (D - 1) (D - 1) = 0
⇒ (D² + 1 - 2D) (D - 1) = 0
⇒ D³ - 3D² + 3D - 1 = 0 ----(1)
The standard Cauchy’s differential equation is in the form of
K₁ D(D - 1) (D - 2) + K₂ D(D - 1) + K₃ D + K₄ = 0
⇒ K₁ [D³ - 3D² + 2D] + K₂ [D² - D] + K₃D + K₄ = 0 ----(2)
By comparing the above two equations,
K₁ = 1, K₄ = -1
-2K₁ + K₂ = -3 ⇒ K₂ = 0
⇒ K₃ = 1
Now the equation (2) becomes,
D(D - 1) (D - 2) + D - 1 = 0
⇒ D(D - 1) (D - 2) y + Dy - y = 0
\( \Rightarrow {x^3}\frac{{{d^3}y}}{{d{x^3}}} + x\frac{{dy}}{{dx}} - y = 0\)
⇒ x³y''' + xy' - y = 0
Question 6
1 / -0

Consider the equation \(\frac{{dy}}{{dx}} = \frac{{{y^2}}}{x}\) with the boundary y(1) = 1. Which one of the following is the correct range of x for which y is real and finite?

A
\(\left( { - \infty \le x \le - 3} \right) \cup \left( {3 \le x \le \infty } \right)\)

B
-3 ≤ x ≤ 0

C
0 ≤ x ≤ 3

D
\(3 \le x \le \infty \)

Solution

\(\frac{{dy}}{{dx}} = \frac{{{y^2}}}{x}\)
\( \Rightarrow \frac{{dy}}{{{y^2}}} = \frac{{dx}}{x}\)
\( = \smallint \frac{{dy}}{{{y^2}}} = \smallint \frac{{dx}}{x}\;\)
\( \Rightarrow - \frac{1}{{{y}}} = \ln \left( x \right) + c\)
put y(1) = 1
\( \Rightarrow - \frac{1}{{{{\left( 1 \right)}}}} = \ln \left( 1 \right) + c\)
⇒ c = - 1
\(\therefore - \frac{1}{y} = \ln \left( x \right) - 1\)
\( \Rightarrow y = \frac{1}{{1 - \ln \left( x \right)}}\)
For y to be finite
\(1-\;ln \left( x \right) \ne 0\;and\;x > 0\)
⇒ ln (x) ≠ 1
⇒ x ≠ e¹
⇒ x ≠ e
⇒ x ≠ 2.718
So the correct range of x is 0 < x < 2.71 and 2.718 < x ≤ ∞
From the given options, 3 ≤ x ≤ ∞ satisfies the above conditions.
Question 7
1 / -0

Particular integral of \(\left( {{x^2}{D^2} - 2} \right)y = {x^2} + \frac{1}{x}\)

A
\(\frac{1}{3}\left( {{x^2} - \frac{1}{x}} \right)\)

B
\(\frac{1}{3}\left( {x - \frac{1}{x}} \right)\)

C
\(\frac{{ln\;x}}{3}\left( {x^2 - \frac{1}{x}} \right)\)

D
\(\frac{{ln\;x}}{3}\left( {x - \frac{1}{x}} \right)\)

Solution

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:
\(\left( {{x^2}{D^2} - 2} \right)y = {x^2} + \frac{1}{x}\)
\( \Rightarrow \left( {D\left( {D - 1} \right) - 2} \right)y = {e^{2t}} + \frac{1}{{{e^t}}}\)
⇒ (D² – D - 2)y = e^2t + e^-t
\(PI = \frac{1}{{\left( {{D^2}D - 2} \right)}}\left( {{e^{2t}} + {e^{ - t}}} \right)\)
\( = \frac{1}{{\left( {{D^2} - D - 2\;} \right)}}{e^{2t}} + \frac{1}{{\left( {{D^2} - D - 2} \right)}}{e^{ - t}}\)
\( = \frac{t}{{2D - 1}}{e^{2t}} + \frac{t}{{2D - 1}}{e^{ - t}}\)
\( = \frac{t}{3}{e^{2t}} + \frac{t}{{ - 3}}{e^{ - t}}\)
\( = \frac{t}{3}\left( {{e^{2t}} - {e^{ - t}}} \right) = \frac{{ln\;x}}{3}\left( {{x^2} - \frac{1}{x}} \right)\)
Question 8
1 / -0

The complementary solution of the differential equation \({x^2}\frac{{{d^3}y}}{{d{x^3}}} - 4x\frac{{{d^2}y}}{{d{x^2}}} + 6\frac{{dy}}{{dx}} = 4\) is

A
C₁ + C₂ e^-3x + C₃e^-4x

B
C₁ + C₂ e^-3x + C₃e^4x

C
C₁ + C₂ x³ + C₃x⁴

D
C₁ + C₂ e^3x + C₃e^-4x

Solution

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 + C₃e^4t
CF = C1 + C2 x³ + C₃x⁴

Question 9

1 / -0

General solution of the Cauchy-Euler equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 7x\frac{{dy}}{{dx}} + 16y = 0\) is

y = c₁x² + c₂x⁴

y = c₁x² + c₂x^-4

y = (c₁ + c₂ ln x) x⁴

y = c₁x² + c₂x^-4 ln x

Solution

Concept:

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

Roots of Auxiliary Equation	Complementary Function
m_1,m₂, m₃, … (real and different roots)	\({C_1}{e^{{m_1}x}} + {C_2}{e^{{m_2}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m_1,m₁, m₃, … (two real and equal roots)	\(\left( {{C_1} + {C_2}x} \right){e^{{m_1}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m_1,m₁, m₁, m₄… (three real and equal roots)	\(\left( {{C_1} + {C_2}x + {C_3}{x^2}} \right){e^{{m_1}x}} + {C_4}{e^{{m_4}x}} + \ldots\)
α + i β, α – i β, m₃, … (a pair of imaginary roots)	\({e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {C_3}{e^{{m_3}x}} + \ldots\)
α ± i β, α ± i β, m₅, … (two pairs of equal imaginary roots)	\({e^{\alpha x}}\left( {\left( {{C_1} + {C_2}x} \right)\cos \beta x + \left( {{C_3} + {C_4}x} \right)\sin \beta x} \right) + {C_5}{e^{{m_5}x}} + \ldots\)

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

⇒ D²y – 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

⇒ y(x) = (c₁ + c₂ ln x) x⁴

Question 10
1 / -0

The solution of a partial differential equation is in the form of z = f₁ (y - ax) + x f₂ (y - bx).
\(4\frac{{{\partial ^2}z}}{{\partial {x^2}}} + 12\frac{{{\partial ^2}z}}{{\partial x\;\partial y}} + 9\frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0\)
Then the values of a and b are

A
\(\frac{1}{2},\frac{1}{2}\)

B
\(\frac{{ - 3}}{2},\frac{{ - 3}}{2}\)

C
\(\frac{{ - 1}}{2},\frac{{ - 1}}{2}\)

D
\(\frac{3}{2},\frac{3}{2}\)

Solution

The given partial differential equation can be written as
(4D² + 12DD’ + 9D’²) z = 0
Its auxiliary equation is,
4m² + 12m + 9 = 0
\( \Rightarrow m = \frac{{ - 3}}{2},\frac{{ - 3}}{2}\)
The complete solution is,
z = f₁ (y – 1.5 x) + x f₂ (y – 1.5 x)
Therefore, \(a = \frac{3}{2},\;b = \frac{3}{2}\)

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Engineering Mathematics Test 7

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Engineering Mathematics Test 7

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SHARING IS CARING

Question 1

0-0

04-01

07-06

09-02

Solution

Question 2

\(\frac{{\partial u}}{{\partial t}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{\partial u}}{{\partial x}}\)

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} + \frac{{{\partial ^2}u}}{{\partial {x^2}}} = 0\)

Solution

Question 3

\( \pm \sqrt 2 \)

0

\( \pm \;3\sqrt 2 \)

± 3

Solution

Question 4

z = e-x sin y + ex cos y

z = e-x sin 2y + ex cos 2y

z = ex sin y + e-x cos y

z = ex sin 2y + e-x cos 2y

Solution

Question 5

x3y''' + 2x2y'' - xy' + y = 0

x3y''' + xy' - y = 0

x3y''' + 3x2y'' - 2xy' + 2y = 0

x3y''' + 4x2y'' + xy' - y = 0

Solution

Question 6

\(\left( { - \infty \le x \le - 3} \right) \cup \left( {3 \le x \le \infty } \right)\)

-3 ≤ x ≤ 0

0 ≤ x ≤ 3

\(3 \le x \le \infty \)

Solution

Question 7

\(\frac{1}{3}\left( {{x^2} - \frac{1}{x}} \right)\)

\(\frac{1}{3}\left( {x - \frac{1}{x}} \right)\)

\(\frac{{ln\;x}}{3}\left( {x^2 - \frac{1}{x}} \right)\)

\(\frac{{ln\;x}}{3}\left( {x - \frac{1}{x}} \right)\)

Solution

Question 8

C1 + C2 e-3x + C3 e-4x

C1 + C2 e-3x + C3 e4x

C1 + C2 x3 + C3 x4

C1 + C2 e3x + C3 e-4x

Solution

Question 9

y = c1x2 + c2x4

y = c1x2 + c2x-4

y = (c1 + c2 ln x) x4

y = c1x2 + c2x-4 ln x

Solution

Question 10

\(\frac{1}{2},\frac{1}{2}\)

\(\frac{{ - 3}}{2},\frac{{ - 3}}{2}\)

\(\frac{{ - 1}}{2},\frac{{ - 1}}{2}\)

\(\frac{3}{2},\frac{3}{2}\)

Solution

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z = e^-x sin y + e^x cos y

z = e^-x sin 2y + e^x cos 2y

z = e^x sin y + e^-x cos y

z = e^x sin 2y + e^-x cos 2y

x³y''' + 2x²y'' - xy' + y = 0

x³y''' + xy' - y = 0

x³y''' + 3x²y'' - 2xy' + 2y = 0

x³y''' + 4x²y'' + xy' - y = 0

C₁ + C₂ e^-3x + C₃e^-4x

C₁ + C₂ e^-3x + C₃e^4x

C₁ + C₂ x³ + C₃x⁴

C₁ + C₂ e^3x + C₃e^-4x

y = c₁x² + c₂x⁴

y = c₁x² + c₂x^-4

y = (c₁ + c₂ ln x) x⁴

y = c₁x² + c₂x^-4 ln x