Solve all your doubts with Teachoo Black (new monthly pack available now!)

Ex 9.3

Ex 9.3, 1
Deleted for CBSE Board 2023 Exams

Ex 9.3, 2 Deleted for CBSE Board 2023 Exams

Ex 9.3, 3 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 4 Deleted for CBSE Board 2023 Exams You are here

Ex 9.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 6 Deleted for CBSE Board 2023 Exams

Ex 9.3, 7 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 8 Deleted for CBSE Board 2023 Exams

Ex 9.3, 9 Deleted for CBSE Board 2023 Exams

Ex 9.3, 10 Important Deleted for CBSE Board 2023 Exams

Ex 9.3, 11 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 9.3, 12 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

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