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Question : 160 of 160
Marks:
+1,
-0
Solution:
‌‌+‌=x2‌⇒‌‌‌‌+‌=1‌‌ Put ‌‌‌‌=t⇒−‌‌=‌‌∴‌‌−‌+‌=1⇒‌−‌=−1On comparing with
‌+Py=Q, we get
P=−‌,Q=−1 ‌ I. F. ‌=e∫P‌dx=e−∫‌dy=‌The solution is
‌t‌ (I. F. ) ‌=∫Q‌ (I. F. ) ‌dy+c‌t⋅‌=∫(−1)⋅‌dy+c‌⇒‌‌‌⋅‌=−log‌y+c‌⇒‌‌‌=cy−y‌log‌y‌
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