Test: Mathematical Physics - 1 - Physics MCQ

Comparing with equation standard expression for the metric tensor 

Polynomial, 3x² + x - 1

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given, 

⇒ Degree = 2 and Order = 5.

∴ The order and degree of the differential equation are 2, 5 repectively.

The correct answer is Option D.

C_n = 1/ inπ

Calculation:

GIven, 
Let y = tan z ⇔ dy = sec²z dz

Let u = ln x ⇔ du = dx/x
⇒ 
Integrating both sides, we get:

⇒ z = tan^-1u + C

⇒ tan^-1y = tan-1(ln x) + C

Now, y(1) = 0

⇒ 0 = 0 + C

⇒ C = 0

∴  tan-1y = tan-1(ln x)

⇒ y(x) = ln x

⇒ y(e³) = ln e³ = 3

∴ The value of y(e³) is 3.

The correct answer is Option B.

Formula Used:

The Taylor series expansion of f(x) about c :

or Taylor Series = 

and

Calculation: 

add equations (i) and (ii)

Hence, option (B) is the correct answer.

For eigen values, 

(1-λ)((1-λ)²-1)-(1-λ-1)+1(1-(1-λ))=0

λ³-3λ²=0
λ=0,0,3
For any n×n  matrix having all elements unity eigenvalues are 0,0,0,...,n

e^y(y′ − 1) = ex

⇒ dy/dx = e^x−y + 1

Let x - y = t

1 – dy/dx = dt/dx

So, we can write

⇒ 1 − dt/dx = e^t + 1

⇒ −e^−t dt = dx

⇒ e^−t = x + c

⇒ e^y−x = x + c

1 = 0 + c

⇒ e^y−x = x + 1

at x = 1

⇒ e^y−1 = 2

⇒ y = 1 + log2²

Hence Option (D) is correct.

(2 + x²u) du + (xu²) dx = 0

For exact differential, dP / dx = dQ / du
Q = (2 + x²u), P = (xu²)

Since dP/du = dQ/dx, hence it is an exact differential.

Thus its solution is given by:

∫Pdx + ∫terms in Q(x,y) not containing x du = C

Given differential equation is 

e^-by dy - e^ax dx = 0
Integrating both sides, we get,

ae^-by + be^ax = abc
ae^-by + be^ax = c₁, where c₁ = abc

dy / dx = x²y − 1,y(0) = 1, y(0, 1) = ?
Here, x₀ = 0, y₀ = 1, h = 0.1 so y(0, 1) = y₁ = ?
Here, y^I(x) = x²y - 1
⇒ y^I(0) = -1
y^II(x) = 2xy + x²y^I
⇒ y^II(0) = 0
y^III(x) = 2y + 2xy^I + 2xy^I + x²y^II
⇒ y^III(0) = 2
y^IV(x) = 2y^I + 4(xy^II + y^I) + 2xy^II + x2y^III
⇒ y^IV(0) = -6 and so on......
Hence by Taylor series:

1 - 0.1 + 0 + 0.00033 + ......
= 0.90031 ≈ 0.90033