(a) method of undetermined coefficients
y′′=6x+xex
y=yu+yp1+yp2 ,
yu′′=0 ⟹ λ1,2=0 ⟹ yu=C1x+C2 ,
yp1′′=6x; ⟹yp1′=∫6xdx=3x2; ⟹yp1=∫3x2dx=x3 ,
yp2′′=xex ,
yp2(x)=(Ax+B)ex ,
yp2′(x)=Aex+ (Ax+B)ex=(A+Ax+B)ex ,
yp2′′(x)=Aex+(A+Ax+B)ex=ex(2A+B+Ax)=xex,
A=1;2A+B=0
A=1;B=−2
yp2(x)=(Ax+B)ex=(x−2)ex ,
Answer: y=C1x+C2+x3+(x−2)ex .
(b) method of variation of parameters
y′′=6x+xex=f(x) ,
y=yu+yp ,
yu′′=0 ⟹ λ1,2=0 ⟹ yu=C1x+C2 ,
make up a system of equations:
{C1′y1+C2′y2=0C1′y1′+C2′y2′=f(x) ,
{C1′x+C2′=0C1′+C2′0=6x+xex ,
{xC1′+C2′=0C1′=6x+xex ,
C1(x)=∫(6x+xex)dx=3x2+ex(x−1)+C1 ,
C2=−∫(6x+xex)xdx=−∫(6x2+x2ex)dx=−2x3−ex(x2−2x+2)+C2 ,
y=C1(x)x+C2(x)=(3x2+ex(x−1)+C1)x−2x3−ex(x2−2x+2)+C2=x3+(x−2)ex+C1x+C2
Answer: y=x3+(x−2)ex+C1x+C2 .
NB: ∫xexdx=(dx=duu=x)(v=exdv=exdx)=xex−∫exdx=ex(x−1)+C ,
∫x2exdx=(2xdx=duu=x2)(v=exdv=exdx)=x2ex−2∫xexdx=x2ex−2ex(x−1)+C=ex(x2−2x+2)+C.
Comments
Leave a comment