From y′′+y=sinx, the auxiliary equation is;
m2+1=0⇒m=±i
Thus, the complementary function is;
yc=c1cosx+c2sinx where c1 and c2 are arbitrary constant.
Next, to obtain the particular integral using method of undetermined coefficients.
Let yp=axcosx+bxsinx, then
yp′′=2bcosx−2asinx−axcosx−bxcosx
Substituting yp and yp′′ into the given DE yields;
(2bcosx−2asinx−axcosx−bxcosx)+(axcosx+bxsinx)=sinx⇒2bcosx−2asinx=sinx
comparing yields;
b=0, and a=−21
Thus, yp=axcosx+bxsinx=−2xcosx+0×xsinx=−2xcosx
Hence, the general solution is;
y=yc+yp=c1cosx+c2sinx−2xcosx
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