solve differential 1st order and 1st degree equations xlogxdy/dx+y=xsin2x
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Step-by-step explanation:
Given differential equation is It is a linear differential equation of the type dy/dx+py=Q => y=logx+C/logx which is the solution of the given differential equationRead more on Sarthaks.com - https://www.sarthaks.com/50426/solve-the-following-differential-equation-xlogx-dy-dx-y-2logx
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the answer is the question(im lying)
xlogxdy/dx+y=xsin2x
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