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

Misc 2 - Chapter 9 Class 12 Differential Equations - Part 10
Misc 2 - Chapter 9 Class 12 Differential Equations - Part 11

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.