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Question : 134 of 150
Marks:
+1,
-0
Solution:
‌2‌=‌+‌ . . . (i)
‌‌ Put ‌‌‌‌=v‌⇒‌‌y=vx‌⇒‌‌‌=v+x‌‌Then, Eq. (i) becomes,
‌2v+2x‌=v2+v⇒‌‌2x‌=v2−v⇒2⋅‌=‌⇒∫‌=‌‌∫‌⇒∫(‌−‌)‌dv=‌‌∫‌⇒‌‌log(‌)=‌‌log‌x+log‌C⇒‌‌log‌=log‌√x‌C ⇒‌‌=√xC⇒‌‌=C√x⇒y‌=x+yC√x
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