Misc 2 - Chapter 9 Class 12 Differential Equations - Part 9

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Misc 2 - Chapter 9 Class 12 Differential Equations - Part 10

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Misc 2 - Chapter 9 Class 12 Differential Equations - Part 11

  1. Chapter 9 Class 12 Differential Equations (Term 2)
  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 . (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 10 years. He provides courses for Maths and Science at Teachoo.