particular integral of
D^2-4D-12=sinx×sin2x
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particular integral
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What is the solution of (d^2 +4) y=sin^2x?
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Harish Chandra Rajpoot
Answered May 22, 2018
Notice,
(D2+4)y=sin2x
complementary function (C.F.) is obtained by putting D2+4=0⟹D=±2i
∴C.F.=C1cos2x+C2sin2x
Now, particular integral (P.I.) is found out as follows
P.I.=sin2xD2+4
=1−cos2x2D2+4
=12(1−cos2xD2+4)
=12(1D2+4−cos2xD2+4)
=12(e0⋅xD2+4−xcos2x2D)
=12(102+4−x21D(cos2x))
=12(14−x2sin2x2)
=18−xsin2x8
∴y=CF+PI
y=C1cos2x+C2sin2x+18−xsin2x8
y=C1cos2x+C2sin2x+1−xsin2x8
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Step-by-step explanation:
बी स्क्वेयर - 40 - 12 * बाय इज इक्वल टू साइन एक्स * 5 * 1
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