The given question is;
(D2+3D+2)y=sin(2x+5)+log3
The Auxiliary equation of the given DE in terms of variable m is;
m2+3m+2=0
⇒(m+1)(m+2)=0⇒m=−1,−2
Hence, the complementary function is;
yc=c1e−x+c2e−2x, where c1 and c2 are arbitrary constants.
Next, to calculate the particular integral (yp) we proceed as follows;
yp=D2+3D+21sin2(x+25)+D2+3D+21(log3)=−4+3D+21sin2(x+25)+log3D2+3D+21e0x=3D−21sin2(x+25)+2log3=9D2−43D+2sin2(x+25)+2log3=−403D+2sin2(x+25)+2log3=40−1[3D(sin(2x+5))+2sin(2x+5)]+2log3=40−1[6cos(2x+5)+2sin(2x+5)]+2log3=−203cos(2x+5)−20sin(2x+5)+2log3
Hence, the general solution is;
y=yc+yp=c1e−x+c2e−2x−203cos(2x+5)−20sin(2x+5)+2log3
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