Question 4 - Forming Differential equations - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Forming Differential equations
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams You are here
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 9 Deleted for CBSE Board 2025 Exams
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 (MCQ) Deleted for CBSE Board 2025 Exams
Question 12 (MCQ) Important Deleted for CBSE Board 2025 Exams
Forming Differential equations
Last updated at April 16, 2024 by Teachoo
Question 4 Form a differential equation representing the given family of curves by eliminating arbitrary constants 𝑎 and 𝑏. 𝑦=𝑒^2𝑥 (𝑎+𝑏𝑥) The Number Of Times We Differentiate Is Equal To Number Of Constants 𝑦=𝑒^2𝑥 (𝑎+𝑏𝑥) ∴ Differentiating Both Sides w.r.t. 𝑥 𝑦^′=𝑑/𝑑𝑥 [𝑒^2𝑥 [𝑎+𝑏𝑥]] 𝑦^′=𝑑[𝑒^2𝑥 ]/𝑑𝑥.[𝑎+𝑏𝑥]+𝑒^(2𝑥 ) 𝑑[𝑎 + 𝑏𝑥]/𝑑𝑥 𝑦^′=〖2𝑒〗^2𝑥 [𝑎+𝑏𝑥]+𝑒^2𝑥.𝑏 𝑦^′=𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏] Again differentiating w.r.t.x 𝑦^′=𝑑/𝑑𝑥 (𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏]) y” = (𝑑 (𝑒^2𝑥))/𝑑𝑥 [2𝑎+2𝑏𝑥+𝑏]+𝑒^2𝑥 (𝑑 [2𝑎+2𝑏𝑥+𝑏])/𝑑𝑥 y” = 2𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏]+𝑒^2𝑥×2𝑏 Putting y’=𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏] y” = 2y’ + 𝑒^2𝑥×2𝑏 y” = 2y’ + 2𝑒^2𝑥 𝑏 y” − 2y’ = 2𝑒^2𝑥 𝑏 Also, y’ − 2y = 𝑒^2𝑥 [2𝑎+2𝑏𝑥 +𝑏]−2𝑒^2𝑥 (𝑎+𝑏𝑥) y’ − 2y = 2a𝑒^2𝑥+2𝑏𝑥 𝑒^2𝑥+𝑒^2𝑥 𝑏−2𝑎〖 𝑒〗^2𝑥−2𝑏𝑥 𝑒^2𝑥 y’ − 2y = (2𝑎〖 𝑒〗^2𝑥−2𝑎〖 𝑒〗^2𝑥 )+(2𝑏𝑥 𝑒^2𝑥−2𝑏𝑥 𝑒^2𝑥 )+𝑒^2𝑥 𝑏 y’ − 2y = 0 + 0 + 𝑒^2𝑥 𝑏 y’ − 2y = 𝑒^2𝑥 𝑏 Now ((1))/((2)) , (𝑦" − 2𝑦)/(𝑦^(′ ) − 2𝑦)=(2𝑒^2𝑥 𝑏)/(𝑒^2𝑥 𝑏) (𝑦^′′ − 2𝑦^′)/(𝑦^′−2𝑦)= 2 y” − 2y’ = 2(y’ − 2y) y” − 2y’ = 2y’ − 4y y” − 2y’ − 2y’ + 4y = 0 y” − 4y’ + 4y = 0 (As 𝑦=𝑒^2𝑥 (𝑎+𝑏𝑥) ) Again differentiating w.r.t.x y” =𝑑/𝑑𝑥 (𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏]) y” = (𝑑 (𝑒^2𝑥))/𝑑𝑥 [2𝑎+2𝑏𝑥+𝑏]+𝑒^2𝑥 (𝑑 [2𝑎+2𝑏𝑥+𝑏])/𝑑𝑥 y” = 2𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏]+𝑒^2𝑥×2𝑏 Putting y’=𝑒^2𝑥 [2𝑎+2𝑏𝑥+𝑏] y” = 2y’ + 𝑒^2𝑥×2𝑏 y” = 2y’ + 2𝑒^2𝑥 𝑏 y” − 2y’ = 2𝑒^2𝑥 𝑏 …(1) Differentiating again w.r.t x 𝑦^′′−2𝑦^′=𝑑(𝑒^2𝑥.𝑏)/𝑑𝑥 𝑦^′′−2𝑦^′=2𝑒^2𝑥 𝑏 Dividing (1) and (2) i.e. ((2))/((1)) , (𝑦" − 2𝑦)/(𝑦^(′ ) − 2𝑦)=(2𝑒^2𝑥 𝑏)/(𝑒^2𝑥 𝑏) (𝑦^′′ − 2𝑦^′)/(𝑦^′−2𝑦)= 2 y” − 2y’ = 2(y’ − 2y) …(2) y” − 2y’ = 2y’ − 4y y” − 2y’ − 2y’ + 4y = 0 y” − 4y’ + 4y = 0 is the required equation