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Question : 66 of 160
Marks:
+1,
-0
Solution:
We have,
y=eax(cos‌b‌x+sin‌b‌x) ‌=eax(−sin‌b‌x⋅b+cos‌b‌x⋅b) +a⋅eax(cos‌b‌x+sin‌b‌x) ‌=beax(cos‌b‌x−sin‌b‌x)+ay ⇒‌‌‌=b[eax(−sin‌b‌x⋅b−cos‌b‌x⋅b) +a⋅eax(cos‌b‌x−sin‌b‌x)]+a‌ ⇒‌‌‌=b [−beax(sin‌b‌x+cos‌b‌x)+aeax(cos‌b‌x−sin‌b‌x)] +a‌=0 ⇒‌‌‌=b[(−b)(y)+aeax(cos‌b‌x−sin‌b‌x)]+a‌ ⇒‌‌‌−a‌+b2y−abeax(cos‌b‌x−sin‌b‌x)=0 ⇒‌‌‌−a‌+b2y−a(‌−ay)=0 ⇒‌‌‌−2a‌+(b2+a2)y=0 On comparing with
‌−k‌+Ly=0 ∴‌‌K=2a,L=b2+a2 ∴‌‌L+bk=b2+a2+2ab=(a+b)2
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