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dydxxx 10g x--+ y =-log x |
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Answer» xlogx dy/dx + y = 2/x . logx dividing throughout by xlogx to make it in standard form since it is a linear differential equation dy/dx + y/xlogx = 2logx/x²logx P = 1/xlogx Q = 2/x² integrating factor (IF)= e∫P= e∫1/xlogx= elog(logx)logx solution: y . IF = ∫Q.IF dxylogx = ∫2/x2 . logx dxlet log x = t => x = et1/x dx = dt => ylogx = 2∫t.etdtylogx = 2[t∫etdt -∫(d/dx t . ∫etdt) dt]ylogx = 2[t.et- ∫1.etdt]ylogx = 2[t.et- et] replacing tylogx = 2[logx.elogx- elogx] |
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