Chapter 5We use initial conditions to determine constantsC1,C2,C3,andC4. Diferentiating (5.7) and(5.8), we get3=x(0) =C1e0+C2e−0+C3cos 0 +C4sin 0 =C1+C2+C3,1=x0(0) =C1e0−C2e−0−C3sin 0 +C4cos 0 =C1−C2+C4,1=y(0) =C1e0+C2e−0−C3cos 0−C4sin 0 =C1+C2−C3,−y0(0) =C1e0−C2e−0+C3sin 0−C4cos 0 =C1−C2−C4⇒C1+C2+C3=3,C1−C2+C4=1,C1+C2−C3,C1−C2−C4=−1.Solving we obtainC1=C2=C3=C4= 1. So, the desired solution isx(t)=et+e−t+cost+sint,y(tet+e−t−cost−sint.23.We will attempt to solve this system by Frst eliminating the ±unctiony(t). Thus, we multiplythe Frst equation by (D+ 2) and the second by−(D−1). There±ore, we obtain(D+2)(D−1)[x]+(DD−1)D[y]=(D+2)±−3e−2t²=6e−2t−6e−2t=0,−(D−1)(D+2)[x]−(D−1)(Dy]=−(D−1)±3et²=−3et+3et.Adding these two equations yields
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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.