Chapter 6Diferentiatingyp(x) yieldsy0p(x)=2C1xe−2x−2C1x2e−2x=2C1(x−x2)e−2x⇒y0p(x)=−4C1(x−x2)e−2x+2C1(1−2x)e−2xC1(2x2−4x+1)e−2x⇒y0p(x−4C1(2x2−4xe−2xC1(4x−4)e−2x=4C1(−2x2+6x−3)e−2x.By substituting these expressions into the nonhomogeneous equation, we obtainy0p(x)+3y0p(x)−4yp(x)=4C1(−2x2x−3)e−2x+6C1(2x2−4xe−2x−4C1x2e−2x=e−2x⇒−6C1e−2x=e−2x.By equating coeﬃcients, we see thatC1=−1/6. Thus, a general solution to the nonhomoge-neous diferential equation is given byy(xyh(x)+yp(xc1ex+c2e−2x+c3xe−2x−16x2e−2x.9.Solving the auxiliary equation,r3−3r2+3r−1=(r−1)3= 0, we Fnd thatr= 1 is its rooto± multiplicity three. There±ore, a general solution to the associated homogeneous equation is
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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 Berkeley.