Exercises 6.3By diferentiatingyp(x), we Fndy0p(x)=C1xex+C1ex+C3+2C4x⇒y0p(xC1xexC1exC4⇒y0p(xC1xex+3C1ex.Substituting these expressions into the original diferential equation, we obtainy0p(x)−2y0p(x)−5y0p(x)+6yp(xC1xexC1ex−2C1xex−4C1ex−4C4−5C1xex−5C1ex−5C3−10C4x+6C1xexC2C3xC4x2=ex+x2⇒−6C1ex+(−4C4−5C3C2)+(−10C4C3)xC4x2=ex+x2.Equating coeﬃcients yields−6C1=1⇒C1=−16,6C4⇒C4=16,−10C4C3=0⇒C3=10C46=1036=518,−4C4−5C3C2⇒C2=4C4+5C36=4(1/6) + 5(5/18)6=37108.Thus, a general solution to the nonhomogeneous equation is given byy(xyh(x)+yp(xc1ex+c2e3x+c3e−2x−16xex+16x2+518x+37108.7.In Problem 3, a general solution to the associated homogeneous equation was ±ound to be
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.