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Question Number 196843 by sonukgindia last updated on 01/Sep/23
Answered by qaz last updated on 02/Sep/23
xyy″=yy′+x−x(y′)2⇒x(yy′)′=yy′+xyy′=e∫dxx(C1+∫e−∫dxxdx)=C1x+xlnx⇒y2=C1x2+x2lnx+C2,C1=12,C2=72y2=12x2+x2lnx+72
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