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Ex 9.3, 4 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.3, 4 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.3, 4 - Chapter 9 Class 12 Differential Equations - Part 4

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Ex 9.3, 4 Form a differential equation representing the given family of curves by eliminating arbitrary constants ğ‘Ž and 𝑏. 𝑦=𝑒^2𝑥 (ğ‘Ž+𝑏𝑥) The Number Of Times We Differentiate Is Equal To Number Of Constants 𝑦=𝑒^2𝑥 (ğ‘Ž+𝑏𝑥) ∴ Differentiating Both Sides w.r.t. 𝑥 𝑦^′=𝑑/𝑑𝑥 [𝑒^2𝑥 [ğ‘Ž+𝑏𝑥]] 𝑦^′=𝑑[𝑒^2𝑥 ]/𝑑𝑥.[ğ‘Ž+𝑏𝑥]+𝑒^(2𝑥 ) 𝑑[ğ‘Ž + 𝑏𝑥]/𝑑𝑥 𝑦^′=〖2𝑒〗^2𝑥 [ğ‘Ž+𝑏𝑥]+𝑒^2𝑥.𝑏 𝑦^′=𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏] Again differentiating w.r.t.x 𝑦^′=𝑑/𝑑𝑥 (𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]) y” = (𝑑 (𝑒^2𝑥))/𝑑𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]+𝑒^2𝑥 (𝑑 [2ğ‘Ž+2𝑏𝑥+𝑏])/𝑑𝑥 y” = 2𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]+𝑒^2𝑥×2𝑏 Putting y’=𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏] y” = 2y’ + 𝑒^2𝑥×2𝑏 y” = 2y’ + 2𝑒^2𝑥 𝑏 y” − 2y’ = 2𝑒^2𝑥 𝑏 Also, y’ − 2y = 𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥 +𝑏]−2𝑒^2𝑥 (ğ‘Ž+𝑏𝑥) y’ − 2y = 2a𝑒^2𝑥+2𝑏𝑥 𝑒^2𝑥+𝑒^2𝑥 𝑏−2ğ‘Žã€– 𝑒〗^2𝑥−2𝑏𝑥 𝑒^2𝑥 y’ − 2y = (2ğ‘Žã€– 𝑒〗^2𝑥−2ğ‘Žã€– 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 (ğ‘Ž+𝑏𝑥) ) Again differentiating w.r.t.x y” =𝑑/𝑑𝑥 (𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]) y” = (𝑑 (𝑒^2𝑥))/𝑑𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]+𝑒^2𝑥 (𝑑 [2ğ‘Ž+2𝑏𝑥+𝑏])/𝑑𝑥 y” = 2𝑒^2𝑥 [2ğ‘Ž+2𝑏𝑥+𝑏]+𝑒^2𝑥×2𝑏 Putting y’=𝑒^2𝑥 [2ğ‘Ž+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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.