Ex 9.3, 7 - Chapter 9 Class 12 Differential Equations

Last updated at August 19, 2026 by

Ex 9.3, 7 - Find general solution: y log y dx - x dy = 0 - Ex 9.3

part 2 - Ex 9.3, 7 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Ex 9.3, 7 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations

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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

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