Question 4 - Formation of Differntial equation when general solution given - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Formation of Differntial equation when general solution given
Formation of Differntial equation when general solution given
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