常微分方程,第二题怎么做?
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因为y1(x),y2(x)是y'+P(x)y=Q(x)的解
所以y1'+P(x)y1=Q(x),y2'+P(x)y2=Q(x)……①
因为py1(x)+2qy2(x)是y'+P(x)y=0的解
所以py1'+2qy2'+pP(x)y1+2qP(x)y2=0
p(y1'+P(x)y1)+2q(y2'+P(x)y2)=0
由①得,pQ(x)+2qQ(x)=0
p+2q=0……②
因为py1(x)-qy2(x)是y'+P(x)y=Q(x)的解
所以py1'-qy2'+pP(x)y1-qP(x)y2=Q(x)
p(y1'+P(x)y1)-q(y2'+P(x)y2)=Q(x)
由①得,pQ(x)-qQ(x)=Q(x)
p-q=1……③
由②③得,p=2/3,q=-1/3
所以y1'+P(x)y1=Q(x),y2'+P(x)y2=Q(x)……①
因为py1(x)+2qy2(x)是y'+P(x)y=0的解
所以py1'+2qy2'+pP(x)y1+2qP(x)y2=0
p(y1'+P(x)y1)+2q(y2'+P(x)y2)=0
由①得,pQ(x)+2qQ(x)=0
p+2q=0……②
因为py1(x)-qy2(x)是y'+P(x)y=Q(x)的解
所以py1'-qy2'+pP(x)y1-qP(x)y2=Q(x)
p(y1'+P(x)y1)-q(y2'+P(x)y2)=Q(x)
由①得,pQ(x)-qQ(x)=Q(x)
p-q=1……③
由②③得,p=2/3,q=-1/3
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