Differential Equation

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

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

Answer :

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

                   ∴       degree = 2

Answer :

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

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