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±è½å¤æ ¹ï¼1+2iï¼1--2iï¼å解为e^x(C1cos2x+C2sin2x)ï¼
åæ±éé½æ¬¡æ¹ç¨çç¹è§£ï¼ä»¤y=e^x*f(x)ï¼åy'=e^x(f+f')ï¼yââ=e^x(f+2f'+f'')ï¼ä»£å
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f''+4f=sin2xãç±æ¤å¾ãcos2x*f'(x)+2sin2x*f(x)ã'=cos2x*f''(x)--2sin2x*f'(x)+2sin2x*f'(x)+4cos2x*f(x)
=cos2x*ãsin2xãï¼æ
cos2x*f'(x)+2sin2x*f(x)=--1/4*(cos2x)^2+Cï¼åC=0ã
åå¾(f(x)/cos2x)'=ãcos2x*f'(x)+2sin2x*f(x)ã/(cos2x)^2=--1/4ï¼å æ¤
f(x)/cos2x=--1/4xï¼f(x)=--0.25xcos2xã
æåå¾å解为e^x(C1cos2x+C2sin2x--0.25xcos2x)ã
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ä¸ å°±æ¯æ ¹æ®f''+4fçå½¢å¼èèe^(2ix)(f'ï¼2if)ç导å½æ°ï¼e^(2ix)(f''+4f)ï¼ä½è¿æ ·çè¯ä¼æå¤è¿ç®ï¼å æ¤åªèèå
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