Question

find particular integral of (D²-4D+3)y=sin3xcox3x​

Answer 1

Answer:

How can I solve (D2+4D+3)y=sinx+x∗e3x ?

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(D2+4D+3)y=sinx+x∗e3x(1)

COMPLIMENTARY FUNCTION

Aux. equation m2+4m+3=0⟹m=−1,−3

yc=C1e−x+C2e−3x

PARTICULAR INTEGRAL

yp=1D2+4D+3sinx+1D2+4D+3(x∗e3x)

=1−1+4D+3sinx+(x−2D+4D2+4D+3)1D2+4D+3e3x

=12(1+2D)sinx+(x−2∗3+432+4∗3+3)132+4∗3+3e3x

=1−2D2(1−4D2)sinx+124(x−512)e3x

=sinx−2cosx2(1+4)+124(x−512)e3x

=sinx−2cosx10+(12x−5288)e3x

y=C1e−x+C2e−3x+sinx−2cosx10+(12x−5288)e3x

Alternately ,

yp=1D2+4D+3sinx+1(D+3)(D+1)(x∗e3x)

=sinx−2cosx10+e3x1(D+6)(D+4)(x)

=sinx−2cosx10+e3x1D2+10D+24(x)

=sinx−2cosx10+e3x124(1+D2+10D24)(x)

=sinx−2cosx10+e3x(1+D2+10D24)−124(x)

=sinx−2cosx10+e3x1−D2+10D24+...24(x)

=sinx−2cosx10+e3xx−102424