Chapter 6yields the rootsr=1,−2, and 2. Thus a general solution has the formy(x)=c1ex+c2e−2x+c3e2x.Next, we Fnd constantsc1,c2,andc3such that the solution satisFes the initial conditions.Di±erentiatingy(x) and substituting the initial conditions, we obtain the systemy(0) =(c1ex+c2e−2x+c3e2x)±±x=0=c1+c2+c3=−4,y0(0) =(c1ex−2c2e−2x+2c3e2x)±±x=0=c1−2c2c3=−1,y0(0) =(c1ex+4c2e−2xc3e2x)±±x=0=c1c2c3=−19.Solving yieldsc1=1,c2=−23=−3.With these coeﬃcients, the solution to the given initial problem isy(xex−2e−2x−3e2x.21.By inspection,r= 2 is a root of the characteristic equation,r3−4r2+7r−6 = 0. ²actoringyieldsr3−4r2r−6=(r−2)(r2−2r+3)=0.Therefore, the other two roots are the roots ofr2−2r+3=0,which arer±√2iso a general solution to the given di±erential equation is given by
This is the end of the preview. Sign up
access the rest of the document.
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.