1.The given equation is (D2−2D+5)y=exsinx.
The auxiliary equation is m2−2m+5=0. Solving for m, we get m=1±2i.
Thus, the complementary function is C.F=ex(c1cos2x+c2sin2x).
To find the particular integral, let yp=Aexsinx be the trial solution.
Then,
yp′=Aex(cosx+sinx) and yp′′=A(ex(−sinx+cosx)+ex(cosx+sinx)) =2Aexcosx
Therefore,
2Aexcosx−2(Aex(cosx+sinx))+5Aexsinx=exsinxA=31
Hence the particular integral is yp=31exsinx .
The general solution is y=ex(c1cos2x+c2sin2x)+31exsinx
2.The given equation is (D2−D−2)y=e2x.
The auxiliary equation is m2−m−2=0. Solving for m, we get m=−1,2.
Thus, the complementary function is C.F=c1e−x+c2e2x.
To find the particular integral, since g(x)=e2x is a linear combination of a term in complementary function we choose yp=Axe2x as the trial particular solution.
Thus,
yp′=A(2xex+e2x)=Ae2x(2x+1)yp′′=A(2(2xex+e2x)+2e2x)=4Ae2x(x+1)
Therefore,
4Ae2x(x+1)−Ae2x(2x+1)−2Axe2x=e2xA(4x+4−2x−1−2x)=1A=31
Hence the particular integral is, yp=31xe2x
The general solution is, y=c1e−x+e2x(3x+c2)
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