


Ex 9.3
Ex 9.3, 2 Deleted for CBSE Board 2022 Exams
Ex 9.3, 3 Important Deleted for CBSE Board 2022 Exams
Ex 9.3, 4 Deleted for CBSE Board 2022 Exams You are here
Ex 9.3, 5 Important Deleted for CBSE Board 2022 Exams
Ex 9.3, 6 Deleted for CBSE Board 2022 Exams
Ex 9.3, 7 Important Deleted for CBSE Board 2022 Exams
Ex 9.3, 8 Deleted for CBSE Board 2022 Exams
Ex 9.3, 9 Deleted for CBSE Board 2022 Exams
Ex 9.3, 10 Important Deleted for CBSE Board 2022 Exams
Ex 9.3, 11 (MCQ) Deleted for CBSE Board 2022 Exams
Ex 9.3, 12 (MCQ) Important Deleted for CBSE Board 2022 Exams
Last updated at Dec. 10, 2019 by Teachoo
Ex 9.3, 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