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

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 e^2t + c₂ e^t

⇒ y = c₁x + c₂x²

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 = e^t

⇒ t = log x

Now the equation becomes,

Auxillary equation is D² + 2D + 1 = 0

⇒ (D + 1)² = 0

⇒ D = -1, -1

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 = e^t

⇒ 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 + c₂ t) e^t

= t + 2

Complete solution is

y = (c₁ + c₂ t) e^t + t + 2

Putting t = log x

⇒ y = (c₁ + c₂ log x) x + log x + 2

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

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 = e^z

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.

Concept:

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

Calculation:

Given:

Put x = e^t

⇒ 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) = (c₁ + c₂ t) e^4t

⇒ y(x) = (c₁ + c₂ ln x) x⁴

We are given DE shown below;

By putting x = e^z 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 = (C₁ + C₂z)e^z

⇒ C⋅F = (C₁ + C₂ log x)x (∵ x = e^z)

& 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 = e^z)

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

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^iθ 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:

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 = e^t

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) e^0t

y = (At + B)

Put t = log x

y = A log x + B

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

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 = e^z and log x = z and hence d/dz = θ

Analysis:

(x²D² - 4xD + 6)y = x²   ---(1)

xD = θ 

x²D² = θ (θ - 1)

substitute in equation (1):

[(θ² - θ) - 4θ + 6]y = e^2z     ---(2)

It becomes linear differential equation with real constants.

f(θ) y = ϕ (z)

f(θ) = θ² - 5θ + 6, ϕ (z) = e^2z

C.F:

θ² - 5θ + 6 = 0

m² - 5m + 6 = 0

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

∴ C.F = C₁e^2z + C₂e^3z

putting θ = 2, f(θ) = 0

hence,

= -ze^2z

∴ y = c₁e^2z + c₂e^3z - ze^2z

putting, x = e^z

∴ y = c₁ x² + c₂ x³ - x²logx²