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Question 107 of 126

Q.The order and degree of the differential equation dxdy+dydx=0\dfrac{dx}{dy}+\dfrac{dy}{dx}=0 are :

(a) 2, degree not defined
(b) 1, 2
(c) 2, 1
(d) 2, 2
Puducherry TnboardTamil Nadu HSC (DGE) Board 2020MCQ· 1mImportance★★★★★
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Clearing the reciprocal derivative dx/dy=1/(dy/dx)dx/dy=1/(dy/dx) shows the equation reduces to 1+(dy/dx)2=01+(dy/dx)^2=0, giving order 11 and degree 22.

  1. The given equation is dxdy+dydx=0\dfrac{dx}{dy}+\dfrac{dy}{dx}=0.
  2. To find order and degree, the equation must first be written as a polynomial in the derivatives (free of fractional or negative powers of derivatives). Currently dxdy=1/(dydx)\dfrac{dx}{dy}=1/\left(\dfrac{dy}{dx}\right) is a reciprocal (negative power) of dydx\dfrac{dy}{dx}, so it is not yet in that form.
  3. Multiply throughout by dydx\dfrac{dy}{dx} to clear this: dxdy⋅dydx+(dydx)2=0\dfrac{dx}{dy}\cdot\dfrac{dy}{dx}+\left(\dfrac{dy}{dx}\right)^2=0.
  4. Since dxdy⋅dydx=1\dfrac{dx}{dy}\cdot\dfrac{dy}{dx}=1, this simplifies to 1+(dydx)2=01+\left(\dfrac{dy}{dx}\right)^2=0. …

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