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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Misc 2 For each of the exercise given below , verify that the given function (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (i) π‘₯𝑦=π‘Ž 𝑒^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 : π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 )+2 𝑑𝑦/𝑑π‘₯βˆ’π‘₯𝑦+π‘₯^2βˆ’2=0 π‘₯𝑦=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 Differentiating w.r.t x (𝑑(π‘₯𝑦))/𝑑π‘₯=𝑑/𝑑π‘₯ [π‘Ž 𝑒^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 ] 𝑑π‘₯/𝑑π‘₯ y+𝑑𝑦/𝑑π‘₯ π‘₯ =π‘Žγ€– 𝑒〗^π‘₯+(βˆ’1)𝑏 𝑒^(βˆ’π‘₯)+2π‘₯ y+y^β€² x =aγ€– eγ€—^xβˆ’b e^(βˆ’x)+2x Differentiating again w.r.t x 𝑦′+(𝑦^β€² π‘₯)^β€² =(π‘Žπ‘’^π‘₯ )^β€²βˆ’(𝑏𝑒^(βˆ’π‘₯) )^β€²+(2π‘₯)^β€² 𝑦^β€²+(𝑦^β€²β€² π‘₯+𝑦^β€²Γ—1)=π‘Žπ‘’^π‘₯+𝑏𝑒^(βˆ’π‘₯)+2 𝑦^β€²β€² π‘₯+2𝑦^β€²=π‘Žπ‘’^π‘₯+𝑏𝑒^(βˆ’π‘₯)+2 Now, we know that π‘₯𝑦=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 π‘₯π‘¦βˆ’π‘₯^2=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯) π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)=π‘₯π‘¦βˆ’π‘₯^2 ...(1) (Given equation) ...(2) Putting (2) in (1) 𝑦^β€²β€² π‘₯+2𝑦^β€²=𝒂𝒆^𝒙+𝒃𝒆^(βˆ’π’™)+2 𝑦^β€²β€² π‘₯+2𝑦^β€²=π’™π’šβˆ’π’™^𝟐+2 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+2 𝑑𝑦/𝑑π‘₯=π’™π’šβˆ’π’™^𝟐+2 (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) 𝒙+𝟐 π’…π’š/π’…π’™βˆ’π’™π’š+𝒙^𝟐=𝟎 ∴ The given function is a solution Misc 2 For each of the exercise given below , verify that the given function (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (ii) 𝑦=𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— ) : (𝑑^2 𝑦)/(𝑑π‘₯^2 )βˆ’2 𝑑𝑦/𝑑π‘₯+2𝑦=0 𝑦=𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— ) : Differentiating w.r.t. x 𝑦^β€²=[𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— )]^β€² 𝑦^β€²=γ€–(𝑒〗^π‘₯)β€² (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— )+𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— )β€² 𝑦^β€²=𝒆^𝒙 (𝒂 𝒄𝒐𝒔⁑〖𝒙+𝒃 π’”π’Šπ’β‘π’™ γ€— )+𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— ) Putting 𝑦=𝑒^π‘₯ (π‘Ž π‘π‘œπ‘ β‘γ€–π‘₯+𝑏 𝑠𝑖𝑛⁑π‘₯ γ€— ) : 𝑦^β€²=π’š+𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— ) 𝑦^β€²βˆ’π‘¦=𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— ) Differentiating again w.r.t x 𝑦^β€²β€²βˆ’π‘¦^β€²=[𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )]^β€² 𝑦^β€²β€²βˆ’π‘¦^β€²=γ€–(𝑒〗^π‘₯)β€²(βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )+𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )β€² 𝑦^β€²β€²βˆ’π‘¦^β€²=𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )+𝑒^π‘₯ (βˆ’π‘Ž cos⁑〖π‘₯βˆ’π‘ si𝑛⁑π‘₯ γ€— ) 𝑦^β€²β€²βˆ’π‘¦^β€²=𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )βˆ’π‘’^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 si𝑛⁑π‘₯ γ€— ) Putting 𝑦=𝑒^π‘₯ (π‘Ž π‘π‘œπ‘ β‘γ€–π‘₯+𝑏 𝑠𝑖𝑛⁑π‘₯ γ€— ) 𝑦^β€²β€²βˆ’π‘¦^β€²=𝑒^π‘₯ (βˆ’π‘Ž sin⁑〖π‘₯+𝑏 cos⁑π‘₯ γ€— )βˆ’π‘¦ Putting 𝑦^β€²βˆ’π‘¦=𝑒^π‘₯ (βˆ’π‘Ž 𝑠𝑖𝑛⁑〖π‘₯+𝑏 π‘π‘œπ‘ β‘π‘₯ γ€— ) from (1) …(1) 𝑦^β€²β€²βˆ’π‘¦^β€²=𝑦^β€²βˆ’π‘¦βˆ’π‘¦ 𝑦^β€²β€²βˆ’π‘¦^β€²=𝑦^β€²βˆ’2𝑦 𝑦^β€²β€²βˆ’π‘¦^β€²βˆ’π‘¦^β€²+2𝑦=0 𝑦^β€²β€²βˆ’2𝑦^β€²+2𝑦=0 (𝑑^2 𝑦)/(𝑑π‘₯^2 )βˆ’2 𝑑𝑦/𝑑π‘₯+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 (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (iii) 𝑦=π‘₯ sin⁑3π‘₯ : (𝑑^2 𝑦)/(𝑑π‘₯^2 )+9π‘¦βˆ’6 cos⁑〖3π‘₯=0γ€— 𝑦=π‘₯ sin⁑3π‘₯ Differentiating w.r.t x 𝑦^β€²=(π‘₯ 𝑠𝑖𝑛⁑3π‘₯ )^β€² 𝑦^β€²=π‘₯^β€² sin⁑3π‘₯+π‘₯(sin⁑3π‘₯)β€² 𝑦^β€²=sin⁑3π‘₯+π‘₯Γ—3 cos⁑3π‘₯ 𝑦^β€²=sin⁑3π‘₯+3π‘₯ cos⁑3π‘₯ Differentiating again w.r.t. x 𝑦^β€²β€²=(sin⁑3π‘₯ )^β€²+(3π‘₯ cos⁑3π‘₯ )^β€² 𝑦^β€²β€²=3 cos⁑3π‘₯+γ€–3(π‘₯)γ€—^β€² cos⁑3π‘₯+3π‘₯ (cos⁑3π‘₯ )^β€² 𝑦^β€²β€²=3 cos⁑3π‘₯+3 cos⁑3π‘₯+3π‘₯(βˆ’3 sin⁑3π‘₯) 𝑦^β€²β€²=6 cos⁑3π‘₯βˆ’9π‘₯ sin⁑3π‘₯ Putting 𝑦=π‘₯ sin⁑3π‘₯ 𝑦^β€²β€²=6 cos⁑3π‘₯βˆ’9𝑦 𝑦^β€²β€²+9π‘¦βˆ’6 cos⁑3π‘₯=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 (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (iv) π‘₯^2=2𝑦^2 log⁑𝑦 : (π‘₯^2+𝑦^2 ) 𝑑𝑦/𝑑π‘₯βˆ’π‘₯𝑦=0 π‘₯^2=2𝑦^2 log⁑𝑦 Differentiating Both sides w.r.t. x (π‘₯^2 )^β€²=(2𝑦^2 log⁑𝑦 )β€² 2π‘₯=(2𝑦^2 )^β€² log⁑𝑦 +2𝑦^2 (log⁑𝑦 )^β€² 2π‘₯=2Γ—2𝑦𝑦^β€² log⁑𝑦 + 2𝑦^2Γ— 1/𝑦 𝑦′ 2π‘₯=4𝑦𝑦^β€² log⁑𝑦 + 2𝑦𝑦′ π‘₯=2𝑦𝑦^β€² log⁑𝑦 + 𝑦𝑦′ π‘₯=〖𝑦𝑦〗^β€² (2 log⁑𝑦+1) Now, from our equation π‘₯^2=2𝑦^2 log⁑𝑦 π‘₯^2/(2𝑦^2 )=log⁑𝑦 π‘™π‘œπ‘”β‘π‘¦=π‘₯^2/(2𝑦^2 ) Putting value of π‘™π‘œπ‘”β‘π‘¦ in (1) π‘₯=〖𝑦𝑦〗^β€² (2 π’π’π’ˆβ‘π’š+1) π‘₯=〖𝑦𝑦〗^β€² (2Γ—π‘₯^2/(2𝑦^2 ) " " +1) π‘₯=〖𝑦𝑦〗^β€² (π‘₯^2/𝑦^2 " " +1) π‘₯=〖𝑦𝑦〗^β€² ((π‘₯^2 + 𝑦^2)/𝑦^2 ) π‘₯=𝑦^β€² ((π‘₯^2 + 𝑦^2)/𝑦) π‘₯𝑦=𝑦^β€² (π‘₯^2 + 𝑦^2 ) 𝑦^β€² (π‘₯^2 + 𝑦^2 )βˆ’π‘₯𝑦=0 Thus, Given Function is a solution of the Differential 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 9 years. He provides courses for Maths and Science at Teachoo.