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NDA 2 2024 Maths, question 90

Topic: Differential equations

Passage · Q89 to Q90

Let ydx+(x−y3)dy=0y dx+(x-y^{3}) dy=0 be a differential equation.

Q90

What is the solution of the differential equation?

  1. (a) y4+2x=cy^{4}+2x=c
  2. (b) y4+3x=cy^{4}+3x=c
  3. (c) 2xy4+x=c2xy^{4}+x=c
  4. (d) 4xy−y4=c4xy-y^{4}=cCorrect answer

Solution.

Answer: (d) 4xy−y4=c4xy-y^{4}=c

  1. Use the product rule y dx+x dy=d(xy).
  2. The equation is d(xy)−y³dy=0.
  3. Integrating gives xy−y⁴/4=C, or 4xy−y⁴=c.
  4. Differentiating this expression recovers four times the original differential equation.