Question 12 (MCQ) - Gen and Particular Solution - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Gen and Particular Solution
Gen and Particular Solution
Last updated at April 16, 2024 by Teachoo
Question 12 Which of the following differential equations has 𝑦=𝑥 as one of its particular solution ? (A) (𝑑^2 𝑦)/(𝑑𝑥^2 )−𝑥^2 𝑑𝑦/𝑑𝑥+𝑥𝑦=𝑥 (B) (𝑑^2 𝑦)/(𝑑𝑥^2 )+𝑥 𝑑𝑦/𝑑𝑥+𝑥𝑦=𝑥 (C) ) (𝑑^2 𝑦)/(𝑑𝑥^2 )−𝑥^2 𝑑𝑦/𝑑𝑥+𝑥𝑦=0 (D) (𝑑^2 𝑦)/(𝑑𝑥^2 )+𝑥 𝑑𝑦/𝑑𝑥+𝑥𝑦=0 𝑦=𝑥 Differentiating both sides w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=1 Again differentiating both sides w.r.t. 𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 )=0 Let us check each Options Option A (𝑑^2 𝑦)/(𝑑𝑥^2 ) −𝑥^2 𝑑𝑦/𝑑𝑥 +𝑥𝑦=𝑥 Putting (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 1, 𝑑𝑦/𝑑𝑥 = 0, y = x 0−𝑥^2 (1)+𝑥(𝑥)=𝑥 −𝑥^2+𝑥^2=𝑥 0=𝑥 Since this is not true ∴ Option (A) is not possible Option B (𝑑^2 𝑦)/(𝑑𝑥^2 ) +𝑥 𝑑𝑦/𝑑𝑥+𝑥𝑦=𝑥 Putting (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 1, 𝑑𝑦/𝑑𝑥 = 0, y = x 0+𝑥(1)+𝑥(𝑥)=𝑥 𝑥+𝑥^2=𝑥 𝑥^2=0 Since this is not true ∴ Option (B) is not possible Option C (𝑑^2 𝑦)/(𝑑𝑥^2 )−𝑥^2 𝑑𝑦/𝑑𝑥+𝑥𝑦=0 Putting (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 1, 𝑑𝑦/𝑑𝑥 = 0, y = x 0−𝑥^2 (1)+𝑥(𝑥)=0 −𝑥^2+𝑥^2=0 0=0 Since this is true ∴ Option (C) is possible Option D (𝑑^2 𝑦)/(𝑑𝑥^2 ) +𝑥 𝑑𝑦/𝑑𝑥+𝑥𝑦=0 Putting (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 1, 𝑑𝑦/𝑑𝑥 = 0, y = x 0+𝑥(1)+𝑥(𝑥)=0 𝑥+𝑥^2=0 𝑥=−𝑥^2 Since this is not true ∴ Option (D) is not possible Thus, Option C is correct