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Test: Cauchy's Equation - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mathematics - Test: Cauchy's Equation

Test: Cauchy's Equation for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Cauchy's Equation questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Cauchy's Equation MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Cauchy's Equation below.
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Test: Cauchy's Equation - Question 1

The solution of differential equationwill be ___________, where c1 and c2 are arbitrary constants.

Detailed Solution for Test: Cauchy's Equation - Question 1

Concept:

Linear differential equations with variable coefficients can be reduced to linear differential equations with constant coefficients by suitable substitutions.

Euler Cauchy Homogeneous linear equation:

Calculation:

Given:

Euler Cauchy equation:

Now, the above differential equation becomes

D (D – 1) y – 2 D y + 2 y = 0

⇒ D2 y – D y – 2 D y + 2 y = 0

⇒ (D2 – 3 D + 2) y = 0

Auxiliary equation:

(D2 – 3 D + 2) = 0

⇒ (D - 2) (D - 1) = 0

⇒ D = 2, 1

Now the roots are real and different, and the general solution of given equation is 

y = c1 e2t + c2 et

⇒ y = c1x + c2x2

Test: Cauchy's Equation - Question 2

The complementary function of the differential equation

Detailed Solution for Test: Cauchy's Equation - Question 2

Concept:

Linear differential equations with variable coefficients can be reduced to linear differential equations with constant coefficients by suitable substitutions.

Euler Cauchy Homogeneous linear equation:

Calculation:

Given:

Euler Cauchy equation:

Put x = et

⇒ t = log x

Now the equation becomes,

Auxillary equation is D2 + 2D + 1 = 0

⇒ (D + 1)2 = 0

⇒ D = -1, -1

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Test: Cauchy's Equation - Question 3

Solve

Detailed Solution for Test: Cauchy's Equation - Question 3

Explanation:

Linear differential equations with variable coefficients which can be reduced to linear differential equations with constant coefficients by suitable substitution.

Euler Cauchy's homogeneous linear equation:

This is a Cauchy’s homogenous linear equation

Put x = et

⇒ t = log x

Now, the equation becomes,

(D(D − 1) − D + 1)y = t ⇒ (D − 1)2y = t

A.E. is (D – 1)2 = 0

⇒ D = 1, 1

C.F. = (c1 + ct) et

= t + 2

Complete solution is

y = (c1 + c2 t) et + t + 2

Putting t = log x

⇒ y = (c1 + c2 log x) x + log x + 2

Test: Cauchy's Equation - Question 4

Every Cauchy sequence is

Detailed Solution for Test: Cauchy's Equation - Question 4

Basic properties of Cauchy sequences:

(i) Every convergent sequence is a Cauchy sequence,

(ii) Every Cauchy sequence of real (or complex) numbers is bounded,

(iii) If in a metric space, a Cauchy sequence possessing a convergent subsequence with limit is itself convergent and has the same limit.

∴ Every Cauchy sequence is bounded

Test: Cauchy's Equation - Question 5

Cauchy’s linear differential equation  can be reduced to a linear differential equation with constant coefficient by using substitution

Detailed Solution for Test: Cauchy's Equation - Question 5

Concept:

Any linear equation of the following form:

is considered as Cauchy’s differential equation. The equation has variable coefficients so its solution becomes tedious but we can convert the above equation into the linear differential equation with constant coefficients

By taking,

log x = z or x = ez

Proof:

log x = z

Taking differentiation on both sides we get,



Now, it can be solved by finding C.F and P.I just like we solve linear differential equations with constant coefficients.

Test: Cauchy's Equation - Question 6

General solution of the Cauchy-Euler equation

Detailed Solution for Test: Cauchy's Equation - Question 6

Concept:

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

Calculation:

Given:

Put x = et

⇒ t = ln x

Now, the above differential equation becomes

D(D – 1)y – 7Dy + 16y = 0

⇒ D2y – Dy – 7Dy + 16y = 0

⇒ (D2 – 8D + 16)y = 0

Auxiliary equation:

(D2 – 8 D + 16) = 0

⇒ D = 4

The solutions for the above roots of auxiliary equations are:

y(t) = (c+ c2 t) e4t

⇒ y(x) = (c1 + c2 ln x) x4

Test: Cauchy's Equation - Question 7

The solution of the differential equation

Detailed Solution for Test: Cauchy's Equation - Question 7

We are given DE shown below;

By putting x = ez we can replace
⇒ (D2 – 2D + 1)y = z (Here F(D) = (D - 1)2

⇒ F(D) = (D - 1)2 = (1 - D)2 has two equal roots Q = 1, 1

So its C⋅F = (C+ C2z)ez

⇒ C⋅F = (C1 + C2 log x)x (∵ x = ez)

& its P⋅I = [F(D)]-1⋅z = (1 - D)-2⋅ z

Since expansion of (1 - D)-2 = (1 + 2D + 3D2 + 4D3 + …)

⇒ P⋅I = [1 + 2D + 3D2 + 4D3 + …](z)

P⋅I = (1 + 2D + 3D2 + 4D3 + …)(z)

⇒ P⋅I = z + 2

⇒ P⋅I = log x + 2 (since x = ez)

So net solution of QE will be P⋅I + C⋅F

⇒ y = C⋅F + P⋅I = (C1 + C2 log x)x + log x + 2

⇒ y = (C1 + C2 log x) x + log x + 2

Test: Cauchy's Equation - Question 8

If f(Z) is an analytical function and (r, θ) denotes the polar co-ordinates, then:

Detailed Solution for Test: Cauchy's Equation - Question 8

Cauchy Riemann Equation in Polar Form:

A function f(z) which is single-valued and possesses a unique derivative with respect to z at all points of a region R, is called an analytic function of z in that region.

If f = u + iv is differentiable at z = re then the Polar Cauchy Riemann equations at (r, θ) and x = rcos θ and y = rsin θ is given by, 

Important Point:

Cauchy Riemann Equation in Rectangular Form:

If f(z) = u (x, y) + iv (x,y) is differentiable at z = x + iy. Then at z the first order patial derivatives of u and v exist and satisfy:

Test: Cauchy's Equation - Question 9

Solve

Which of the following is the solution of above equation, when x = 0?

Detailed Solution for Test: Cauchy's Equation - Question 9

Concept:

The given equation is multiplied with x to make it a equation of Cauchy Euler type.

The given equation is of Cauchy Euler type. Cauchy Euler equation is homogenous linear equation with variable coefficient.

They can be reduced to linear equations with constant coefficients

Calculation:

Given:


Put x = et

log x = t

The given equation is written as

[x2D2 + xD]y = 0      ---(1)

Put xD = D’, x2D2 = D’(D’ - 1)

[D’(D’ - 1) + D’]y = 0

D’2 y = 0

The auxiliary equation is D’2 = 0

The roots of this equation are 0, 0

C.F = (At + B) e0t

y = (At + B)

Put t = log x

y = A log x + B

At x = 0; y is undefined as log 0 is undefined.

Test: Cauchy's Equation - Question 10

Solve: (x2D2 - 4xD + 6)y = x2
where D = d/dx 

Detailed Solution for Test: Cauchy's Equation - Question 10

Cauchy's Equation MCQ Question 2 Detailed Solution

Concept:

If the derivation power and the variable power are the same, then the equation is Cauchy - Euler equation. 

Now,

It can be solved by putting x = ez and log x = z and hence d/dz = θ

Analysis:

(x2D2 - 4xD + 6)y = x2   ---(1)

xD = θ 

x2D= θ (θ - 1)

substitute in equation (1):

[(θ2 - θ) - 4θ + 6]y = e2z     ---(2)

It becomes linear differential equation with real constants.

f(θ) y = ϕ (z)

f(θ) = θ2 - 5θ + 6, ϕ (z) = e2z

C.F:

θ2 - 5θ + 6 = 0

m2 - 5m + 6 = 0

(m - 2) (m - 3) = 0

∴ C.F = C1e2z + C2e3z

putting θ = 2, f(θ) = 0

hence,

= -ze2z

∴ y = c1e2z + c2e3z - ze2z

putting, x = ez

∴ y = cx2 + cx3 - x2logx2 

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