Misc 2 - Chapter 9 Class 12 Differential Equations (Term 2)

Transcript

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 ...(1) (Given equation) ...(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 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 . (ii) 𝑦=𝑒^𝑥 (𝑎 cos⁡〖𝑥+𝑏 sin⁡𝑥 〗 ) : (𝑑^2 𝑦)/(𝑑𝑥^2 )−2 𝑑𝑦/𝑑𝑥+2𝑦=0 𝑦=𝑒^𝑥 (𝑎 cos⁡〖𝑥+𝑏 sin⁡𝑥 〗 ) : Differentiating w.r.t. x 𝑦^′=[𝑒^𝑥 (𝑎 cos⁡〖𝑥+𝑏 sin⁡𝑥 〗 )]^′ 𝑦^′=〖(𝑒〗^𝑥)′ (𝑎 cos⁡〖𝑥+𝑏 sin⁡𝑥 〗 )+𝑒^𝑥 (𝑎 cos⁡〖𝑥+𝑏 sin⁡𝑥 〗 )′ 𝑦^′=𝒆^𝒙 (𝒂 𝒄𝒐𝒔⁡〖𝒙+𝒃 𝒔𝒊𝒏⁡𝒙 〗 )+𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 ) Putting 𝑦=𝑒^𝑥 (𝑎 𝑐𝑜𝑠⁡〖𝑥+𝑏 𝑠𝑖𝑛⁡𝑥 〗 ) : 𝑦^′=𝒚+𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 ) 𝑦^′−𝑦=𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 ) Differentiating again w.r.t x 𝑦^′′−𝑦^′=[𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )]^′ 𝑦^′′−𝑦^′=〖(𝑒〗^𝑥)′(−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )+𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )′ 𝑦^′′−𝑦^′=𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )+𝑒^𝑥 (−𝑎 cos⁡〖𝑥−𝑏 si𝑛⁡𝑥 〗 ) 𝑦^′′−𝑦^′=𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )−𝑒^𝑥 (𝑎 cos⁡〖𝑥+𝑏 si𝑛⁡𝑥 〗 ) Putting 𝑦=𝑒^𝑥 (𝑎 𝑐𝑜𝑠⁡〖𝑥+𝑏 𝑠𝑖𝑛⁡𝑥 〗 ) 𝑦^′′−𝑦^′=𝑒^𝑥 (−𝑎 sin⁡〖𝑥+𝑏 cos⁡𝑥 〗 )−𝑦 Putting 𝑦^′−𝑦=𝑒^𝑥 (−𝑎 𝑠𝑖𝑛⁡〖𝑥+𝑏 𝑐𝑜𝑠⁡𝑥 〗 ) from (1) …(1) 𝑦^′′−𝑦^′=𝑦^′−𝑦−𝑦 𝑦^′′−𝑦^′=𝑦^′−2𝑦 𝑦^′′−𝑦^′−𝑦^′+2𝑦=0 𝑦^′′−2𝑦^′+2𝑦=0 (𝑑^2 𝑦)/(𝑑𝑥^2 )−2 𝑑𝑦/𝑑𝑥+2𝑦=0 Thus, Given Function is a solution of the Differential Equation 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 . (iii) 𝑦=𝑥 sin⁡3𝑥 : (𝑑^2 𝑦)/(𝑑𝑥^2 )+9𝑦−6 cos⁡〖3𝑥=0〗 𝑦=𝑥 sin⁡3𝑥 Differentiating w.r.t x 𝑦^′=(𝑥 𝑠𝑖𝑛⁡3𝑥 )^′ 𝑦^′=𝑥^′ sin⁡3𝑥+𝑥(sin⁡3𝑥)′ 𝑦^′=sin⁡3𝑥+𝑥×3 cos⁡3𝑥 𝑦^′=sin⁡3𝑥+3𝑥 cos⁡3𝑥 Differentiating again w.r.t. x 𝑦^′′=(sin⁡3𝑥 )^′+(3𝑥 cos⁡3𝑥 )^′ 𝑦^′′=3 cos⁡3𝑥+〖3(𝑥)〗^′ cos⁡3𝑥+3𝑥 (cos⁡3𝑥 )^′ 𝑦^′′=3 cos⁡3𝑥+3 cos⁡3𝑥+3𝑥(−3 sin⁡3𝑥) 𝑦^′′=6 cos⁡3𝑥−9𝑥 sin⁡3𝑥 Putting 𝑦=𝑥 sin⁡3𝑥 𝑦^′′=6 cos⁡3𝑥−9𝑦 𝑦^′′+9𝑦−6 cos⁡3𝑥=0 Thus, Given Function is a solution of the Differential Equation 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 . (iv) 𝑥^2=2𝑦^2 log⁡𝑦 : (𝑥^2+𝑦^2 ) 𝑑𝑦/𝑑𝑥−𝑥𝑦=0 𝑥^2=2𝑦^2 log⁡𝑦 Differentiating Both sides w.r.t. x (𝑥^2 )^′=(2𝑦^2 log⁡𝑦 )′ 2𝑥=(2𝑦^2 )^′ log⁡𝑦 +2𝑦^2 (log⁡𝑦 )^′ 2𝑥=2×2𝑦𝑦^′ log⁡𝑦 + 2𝑦^2× 1/𝑦 𝑦′ 2𝑥=4𝑦𝑦^′ log⁡𝑦 + 2𝑦𝑦′ 𝑥=2𝑦𝑦^′ log⁡𝑦 + 𝑦𝑦′ 𝑥=〖𝑦𝑦〗^′ (2 log⁡𝑦+1) Now, from our equation 𝑥^2=2𝑦^2 log⁡𝑦 𝑥^2/(2𝑦^2 )=log⁡𝑦 𝑙𝑜𝑔⁡𝑦=𝑥^2/(2𝑦^2 ) Putting value of 𝑙𝑜𝑔⁡𝑦 in (1) 𝑥=〖𝑦𝑦〗^′ (2 𝒍𝒐𝒈⁡𝒚+1) 𝑥=〖𝑦𝑦〗^′ (2×𝑥^2/(2𝑦^2 ) " " +1) 𝑥=〖𝑦𝑦〗^′ (𝑥^2/𝑦^2 " " +1) 𝑥=〖𝑦𝑦〗^′ ((𝑥^2 + 𝑦^2)/𝑦^2 ) 𝑥=𝑦^′ ((𝑥^2 + 𝑦^2)/𝑦) 𝑥𝑦=𝑦^′ (𝑥^2 + 𝑦^2 ) 𝑦^′ (𝑥^2 + 𝑦^2 )−𝑥𝑦=0 Thus, Given Function is a solution of the Differential Equation