Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i) You are here
Misc 2 (ii) Important
Misc 2 (iii)
Misc 2 (iv) Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11
Misc 12 Important
Misc 13 (MCQ)
Misc 14 (MCQ) Important
Misc 15 (MCQ)
Question 1
Question 2 Important
Question 3 Important
Last updated at Dec. 16, 2024 by Teachoo
Misc 2 For each of the exercise given below , verify that the given function (𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡) is a solution of the corresponding differential equation . (i) 𝑥𝑦=𝑎 𝑒^𝑥+𝑏 𝑒^(−𝑥)+𝑥^2 : 𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 )+2 𝑑𝑦/𝑑𝑥−𝑥𝑦+𝑥^2−2=0 𝑥𝑦=𝑎𝑒^𝑥+𝑏 𝑒^(−𝑥)+𝑥^2 Differentiating w.r.t x (𝑑(𝑥𝑦))/𝑑𝑥=𝑑/𝑑𝑥 [𝑎 𝑒^𝑥+𝑏 𝑒^(−𝑥)+𝑥^2 ] 𝑑𝑥/𝑑𝑥 y+𝑑𝑦/𝑑𝑥 𝑥 =𝑎〖 𝑒〗^𝑥+(−1)𝑏 𝑒^(−𝑥)+2𝑥 𝒚+𝒚^′ 𝒙 =𝒂〖 𝒆〗^𝒙−𝒃 𝒆^(−𝒙)+𝟐𝒙 Differentiating again w.r.t x 𝑦′+(𝑦^′ 𝑥)^′ =(𝑎𝑒^𝑥 )^′−(𝑏𝑒^(−𝑥) )^′+(2𝑥)^′ 𝑦^′+(𝑦^′′ 𝑥+𝑦^′×1)=𝑎𝑒^𝑥+𝑏𝑒^(−𝑥)+2 𝒚^′′ 𝒙+𝟐𝒚^′=𝒂𝒆^𝒙+𝒃𝒆^(−𝒙)+𝟐 Now, we know that 𝑥𝑦=𝑎𝑒^𝑥+𝑏 𝑒^(−𝑥)+𝑥^2 𝑥𝑦−𝑥^2=𝑎𝑒^𝑥+𝑏 𝑒^(−𝑥) 𝒂𝒆^𝒙+𝒃 𝒆^(−𝒙)=𝒙𝒚−𝒙^𝟐 Putting (2) in (1) 𝑦^′′ 𝑥+2𝑦^′=𝒂𝒆^𝒙+𝒃𝒆^(−𝒙)+2 𝑦^′′ 𝑥+2𝑦^′=𝒙𝒚−𝒙^𝟐+2 (𝑑^2 𝑦)/(𝑑𝑥^2 ) 𝑥+2 𝑑𝑦/𝑑𝑥=𝒙𝒚−𝒙^𝟐+2 (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) 𝒙+𝟐 𝒅𝒚/𝒅𝒙−𝒙𝒚+𝒙^𝟐=𝟎 ∴ The given function is a solution