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Ex 9.4, 7 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 20, 2021 by

Ex 9.4, 7 - Find general solution: y log y dx - x dy = 0

Ex 9.4, 7 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 7 - Chapter 9 Class 12 Differential Equations - Part 3

Transcript

Ex 9.4, 7 For each of the differential equations in Exercises 1 to 10, find the general solution : ๐‘ฆ logโกใ€–๐‘ฆ ๐‘‘๐‘ฅ โˆ’๐‘ฅ ๐‘‘๐‘ฆ=0ใ€— ๐‘ฆ logโกใ€–๐‘ฆ ๐‘‘๐‘ฅ โˆ’๐‘ฅ ๐‘‘๐‘ฆ=0ใ€— ๐‘ฆ logโก๐‘ฆ ๐‘‘๐‘ฅ=๐‘ฅ ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ/๐‘ฅ = ๐‘‘๐‘ฆ/(๐‘ฆ logโก๐‘ฆ ) Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ/(๐‘ฆ logโก๐‘ฆ )= โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅใ€— โˆซ1โ–’๐‘‘๐‘ฆ/(๐‘ฆ logโก๐‘ฆ )=logโกใ€–|๐‘ฅ|ใ€—+๐ถ Putting t = log y dt = 1/๐‘ฆ dy dy = y dt Hence, our equation becomes โˆซ1โ–’(๐‘ฆ ๐‘‘๐‘ก)/(๐‘ฆ.๐‘ก)=logโกใ€–|๐‘ฅ|ใ€—+๐ถ โˆซ1โ–’๐‘‘๐‘ก/๐‘ก=logโกใ€–|๐‘ฅ|ใ€—+๐ถ ๐‘™๐‘œ๐‘” |๐‘ก|=๐‘™๐‘œ๐‘”โกใ€–|๐‘ฅ|ใ€—+๐ถ ๐‘™๐‘œ๐‘” |๐‘ก|=๐‘™๐‘œ๐‘”โก|๐‘ฅ|+logโก๐ถ Putting t = log y log (log y) = log x + log c log (log y) = log cx (Using log ab = log ๐‘Ž + log b) Cancelling log log y = cx y = ecx

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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.