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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


Transcript

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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.