Cauchy's Equation MCQ Quiz - Objective Question with Answer for Cauchy's Equation - Download Free PDF

Concept:

• This is a first-order differential equation solvable by substitution.
• Substitute y/x = t to convert it into a variable separable form.
• The equation is solved by integrating both sides and using the initial condition y(1) = 0 to find the constant of integration.
• Final calculation involves evaluating y at x = e and x = e², then computing the required expression.

Calculation:

Given,

x² dy/dx + xy = x² + y²

⇒ dy/dx + y/x = 1 + (y/x)²

Let y/x = t, then y = xt

⇒ dy/dx = t + x dt/dx

Substitute in equation:

⇒ t + x dt/dx + t = 1 + t²

⇒ x dt/dx + 2t = 1 + t²

⇒ x dt/dx = (t − 1)²

⇒ dt/(t − 1)² = dx/x

Integrate both sides:

⇒ ∫ dt/(t − 1)² = ∫ dx/x

⇒ −1/(t − 1) = ln x + C

⇒ −1/(y/x − 1) = ln x + C

⇒ −x/(y − x) = ln x + C

Use y(1) = 0:

⇒ −1/(0 − 1) = ln 1 + C

⇒ 1 = C

So, equation is:

−x/(y − x) = ln x + 1 ....(i)

Put x = e:

−e/(y − e) = 1 + 1

⇒ −e/(y − e) = 2

⇒ −e = 2(y − e)

⇒ e = 2e − 2y

⇒ y = e/2

Put x = e²:

−e²/(y − e²) = ln e² + 1

⇒ −e²/(y − e²) = 2 + 1 = 3

⇒ −e² = 3y − 3e²

⇒ 2e² = 3y

⇒ y = 2e²/3

Find 2(y(e))² / y(e²):

⇒ 2 × (e/2)² ÷ (2e²/3)

⇒ 2 × e²/4 ÷ (2e²/3)

⇒ (e²/2) ÷ (2e²/3)

⇒ (e²/2) × (3/2e²) = 3/4

∴ The required value is 0.75

CONCEPT:

Singular point: A point at which a function f(z) is not analytic is known as a singular point or singularity of the function. 

For example, the function 1/(z-2) has a singular point at z - 2 = 0 i.e. z = 2. 

(we have to make the denominator 0 to find the singular point)

⇒ If z = a is a singularity of f(z) and if there is no other singularity within a small circle surrounding the point z = a then it is said to be an isolated singularity of the function f(z); otherwise it is called non-isolated. 

An isolated singular point is also called a simple pole if the order is 1.

Definition of residue at a pole: Let z = a be simple of a function f(z) 

Then Residue at z = a ⇒ 

Cauchy's residue theorem :

If f(z) is analytic in a closed curve C except at a finite number of poles within C, then 

EXPLANATION:

Given the function. 

f(z) = dw

we find the simple pole by making the denominator 0.

∴ w - z = 0 ⇒ w = z

Applying Cauchy's residue theorem

⇒ f(z) = f(2-i) = 2πi (sum of the residues within contour) = 2πi (1-4i) = 8π + 2πi

Hence the correct answer is option 3.

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:

xD = θ 

x²D² = θ(θ - 1), x = e^z, log x = z

∴ (θ² - θ - θ + 1) y = z

(θ² - 2θ + 1) y = z

Now, 

C.F:

θ² - 2θ + 1 = 0

m² - 2m + 1 = 0

∴ (m - 1)(m - 1) = 0

∴ C.F = e^z(C₁ + C₂z)

= e^z (C₁ + C₂z)

P.I:

=[1 + (θ² - 2θ)]^-1z

We know,

(1 + x)^-1 = 1 - x + x² - x³

∴ P.I = z - θ²z + 2θz

= z + 2θz

Now,

∴ P.I = z + 2

= log x + 2

∴ y = e^z(C₁ + C₂z) + log x + 2

= x(c₁ + c₂ log x) + log x + 2

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 = c1x2 + c2x3 - x2 logx

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) = (c1 + c2 t) e4t

⇒ y(x) = (c1 + c2 ln x) x4

The given equation is homogeneous ordinary differential equation.

x = e^z    

z = ln x

Equation (1) can be written as:

D(D - 1)y – 3Dy + 3y = 0

D² – 4D + 3 = 0

D² – 3D – D + 3 = 0

D(D - 3) – 1 (D - 3) = 0

(D - 1)(D - 3) = 0

D = 1, 3

general solution:

y = C₁e^z + C₂ e^3z

z = ln x

y = C₁ x + C₂ x³      ----(1)

To find C₁ and C₂ the initial condition

y(1) = 1

y(2) = 14

substituting in (1)

C₁ + C₂ = 1

2C₁ + 8C₂ = 14

Solving above equations give

C₁ = -1 and C₂ = 2

y = -x + 2x³

y(1.5) = -1.5 + 2(1.5)³

y(1.5) = 5.25

We are given DE shown below;

By putting x = e^z we can replace  as, (D) (D – 1)y – Dy + y = log e^z = z

⇒ (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)

Where

⇒ 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

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:

Let we have

log x = z

Taking differentiation on both sides we get,

∴

Similarly,

∴

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

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,

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

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:

Take x = e^t ⇒ t = log x

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

Concept:

Euler Cauchy Homogeneous linear equation:

Take x = et ⇒ t = log x

Calculation:

By multiplying with ‘x’ on both sides

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  

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) = (c1 + c2 t) e4t

⇒ y(x) = (c1 + c2 ln x) x4