© examsiri.com
Question : 138 of 150
Marks:
+1,
-0
Solution:
Given differential equation can be written as
‌−(‌)y=cos2xHere,
P=−‌=‌| −sin‌2x |
| cos‌2‌x(‌) |
and
‌‌Q=cos2x∵‌‌mathbb‌F=e∫P‌dx=e−∫‌| 2sin‌2x |
| cos‌2‌x(cos‌2‌x+1) |
‌dxPut
cos‌2‌x=t⇒−2sin‌2x‌dx=dt∴‌‌IF‌=e∫‌(‌)‌dt‌=e∫(‌−‌)‌dt =e[log‌t−log(t+1)]=elog‌=elog‌=‌ Now, solution is
‌y×‌=∫‌×cos2x‌dx+C‌=∫‌×cos2x‌dx+C‌=‌‌∫cos‌2‌x‌dx+C‌=‌×‌+C‌⇒y‌=‌sin‌2x+C‌‌ But ‌y(‌)=‌‌∴‌‌‌×‌=‌sin‌2(‌)+C‌⇒‌‌‌=‌×‌+C‌⇒‌‌‌=‌+C⇒‌‌‌=‌+C⇒C=0From Eq. (i), we get
‌‌=‌sin‌2x+0⇒‌‌y‌=‌‌‌=‌‌‌=‌⋅‌
© examsiri.com
Go to Question: