© examsiri.com
Question : 159 of 200
Marks:
+1,
-0
Solution:
We have,
‌+y=exOn comparing with
‌+Py=QHere,
P=1 and
Q=exNow,
IF=e∫1‌dx=exComplete solution is
⇒‌‌y⋅ex‌=∫ex⋅ex‌dx+C′⇒‌‌yex‌=‌+C′⇒‌‌2yex‌=e2x+C‌[∵2C′=C]
© examsiri.com
Go to Question: