Ex 9.3, 7 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by

Slide15.JPG

Slide16.JPG
Slide17.JPG

Transcript

Ex 9.3, 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’™/๐’™ = ๐’…๐’š/(๐’š ๐ฅ๐จ๐ โก๐’š ) Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ/(๐‘ฆ logโก๐‘ฆ )= โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅใ€— โˆซ1โ–’๐’…๐’š/(๐’š ๐’๐’๐’ˆโก๐’š )=๐ฅ๐จ๐ โกใ€–|๐’™|ใ€—+๐‘ช 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’๐’ˆ |๐’•|=๐’๐’๐’ˆโก|๐’™|+๐ฅ๐จ๐ โก๐‘ช 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

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.