Differential%20Equation

1 )

The degree of the differential equation is

y₂^(3/2) – y₁^(1/2) + 1 = 0

View Solution

Answer :

equation: y₂^(3/2) – y₁^(1/2) + 1 = 0 ⇒ (y₂)^(3/2) = y₁^(1/2) – 1

∴ y₂³ = y₁ – 2y₁^(1/2) + 1

∴ (y₂³ – 1) = y₁ – 2y₁^(1/2)

∴ squaring,

y₂⁶ – 2y₂³ + 1 = y₁² – 4y₁^(3/2) + 4y₁

∴ Here degree = 6

2 )

The order of the differential equation whose general solution is given by y = C₁ e^(x+C)2 + (C₃ + C₄) ∙ sin (x + C₅), where C₁, C₂, C₃, C₄, C₅ are arbitrary Constant is_____.

View Solution

Answer :

y = C₁e⁽C⁾² ∙ e^x + (C₃ + C₄) sin (x + C₅)

y = Ae^x + B sin (x + C) where A, B, C are arbitrary Constant. (3 Constant)

∴ order = 3

3 )

The degree of the differential equation of all curves having normal of constant length c is.

View Solution

Answer :

c = Length of normal = y√[1 + (dx / dy)²] ⇒ y² + y²(dx / dy)² = c²

∴ degree = 2

4 )

The degree of the differential equation (d³y / dx³) + 7(d²y / dx²)³ = x² ∙ log (d²y / dx²) is:

degree doesnâ€™t exist

View Solution

Answer :

Differential equation cannot be expressed in a polynomial form, hence degree does not exists.

5 )

The degree of the differential equation satisfying √(1 + x²) + √(1 + y²) = k[x√(1 + y²) – y√(1 + x²)] is:

View Solution

Answer :

Let x = tan α

y = tan β

Hence equation: √(1 + tan²α) + √(1 + tan²β)

= k[tan α√(1 + tan²β) – tan β√(1 + tan²α)]

∴ sec α + sec β = k(tan α ∙ sec β – tan β ∙ sec α)

∴ [(sec α + sec β) / (tan α ∙ sec β – tan β ∙ sec α)] = k

∴ cot [(α – β) / 2] = k

∴ α – β = 2 cot^–1 (k)

∴ tan^–1x – tan^–1y = 2 cot^–1k

∴ [1 / (1 + x²)] – [1 / (1 + y²)](dy / dx) = 0

Hence degree = 1

Topic Categories