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Question : 78 of 160
Marks:
+1,
-0
Solution:
Given, differential equation is
‌x2(y+1)‌+y2(x+1)2=0,y(1)=2 ‌⇒‌‌x2(y+1)‌=−y2(x+1)2 ‌⇒‌‌‌dy=−‌‌dx Integrating to both sides, we get
∫‌dy=−∫‌‌dx ⇒‌‌∫(‌+‌)dy =−∫(1+‌+‌)‌dx ⇒ln|y|−‌=−(x+2‌ln|x|−‌)+C, where
C= constant
‌⇒ln|y|+2‌ln|x|=‌+‌−x+C ‌⇒ln‌|x2y|=‌+‌−x+C‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) Given,
y(1)=2, so we get
‌⇒‌‌ln‌|12⋅2|=‌+‌−1+C ‌⇒‌‌ln|2|=‌+C⇒c=ln|2|−‌ From, Eq. (i), we get
‌ln‌|x2y|=‌+‌−x+ln|2|−‌ ⇒ln‌|x2y|−ln|2|=‌+‌−x−‌ ⇒ln‌|‌x2y|=‌+‌−x−‌
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