characteristic equation:
k2−3k+2=0
k=23±9−8
k1=1,k2=2
complementary solution:
yc=c1ex+c2e2x
for particular solution:
yp1=Ax2+Bx+C
2A−3(2Ax+B)+2(Ax2+Bx+C)=x2
A=1/2
2A−3B+2C=0
−6A+2B=0⟹B=3/2
C=(3B−2A)/2=(9/2−1)/2=7/4
yp2=Acosx+Bsinx
−Acosx−Bsinx−3(−Asinx+Bcosx)+2(Acosx+Bsinx)=sinx
A−4B=0
3A+B=1
B=1/13,A=4/13
y=yc+yp1+yp2=c1ex+c2e2x+x2/2+3x/2+7/4+4cosx/13+sinx/13
Comments
Leave a comment