â«(x²-e^x)dx=(1/3)x³-e^x+C
â«(x²+sec²x)dx=(1/3)x³+tanx+C
â«[1/(2x-3)]dx=(1/2)â«[1/(2x-3)]d(2x-3)=(1/2)lnâ£2x-3â£+C
â«sin(x/2)dx=2â«sin(x/2)d(x/2)=-2cos(x/2)+C
â«dx/(1+âx)ï¼è®¾âx=uï¼åx=u²ï¼dx=2uduï¼
äºæ¯åå¼=2â«udu/(1+u)=2â«[1-1/(1+u)]du=2[u-lnâ£1+uâ£+C=2[(âx)-ln(1+âx)]+C
â«xe^(3x)dx=(1/3)â«xd[e^(3x)]=(1/3)[xe^(3x)-â«e^(3x)dx]=(1/3)e^(3x)-â«e^(3x)d(3x)
=(1/3)e^(3x)-e^(3x)+C=-(2/3)e^(3x)+C
ã-2ï¼-1ãâ«(1+2x)²dx=ã-2ï¼-1ã(1/2)â«(1+2x)²d(1+2x)=(1/6)(1+2x)³ã-2ï¼-1ã=-(1/6)+9/2=13/3
ã0ï¼1/2ãâ«arcsinxdx=[xarcsinx+â(1-x²)]ã0ï¼1/2ã=Ï/6+(â3/2)-1
ã0ï¼1/4ãâ«xdx/â(1-2x)=ã0ï¼1/4ã-â«xdâ(1-2x)=-[xâ(1-2x)-â«â(1-2x)dx]ã0ï¼1/4ã
=-[xâ(1-2x)+(1/2)â«â(1-2x)d(1-2x)]ã0ï¼1/4ã
=-[xâ(1-2x)+(1/3)xâ(1-2x)+â(1-2x)³]ã0ï¼1/4ã
=-[1/(4â2)+1/(12â2)+1/(2â2)]+1=(12+5â2)/12
(ä¸).æ±å¾®åæ¹ç¨2xy+x²y'=(1-x)yçé解ã
解ï¼x²y'=y-3xyï¼å³æ(3xy-y)dx+x²dy=0...........(1)ï¼
å ¶ä¸P=3xy-yï¼Q=x²ï¼∂P/∂y=3x-1â ∂Q/∂x=2xï¼æ åæ¹ç¨ä¸æ¯å ¨å¾®åæ¹ç¨ã
ä½(1/Q)(∂P/∂y-∂Q/∂x)=(1/x²)(3x-1-2x)=(x-1)/x²=G(x)æ¯xçå½æ°ï¼
æ æ积åå åμ(x)=e^â«G(x)dx=e^â«[(1/x)-(1/x²)]dx=e^(lnx+1/x)=xe^(1/x)ï¼
å°Î¼(x)ä¹(1)å¼ä¸¤è¾¹å¾[(3xy-y)xe^(1/x)]dx+[x³e^(1/x)]dy=0..........(2)
æ¤æ¶P=(3xy-y)xe^(1/x)ï¼Q=x³e^(1/x)ï¼
∂P/∂y=(3x-1)xe^(1/x)=∂Q/∂x=3x²e^(1/x)-xe^(1/x)=(3x-1)xe^(1/x)ï¼æ (2)æ¯å ¨å¾®åæ¹ç¨ã
(2)ç左边æ¯å½æ°u(xï¼y)=â«[(3xy-y)xe^(1/x)]dx=yx³e^(1/x)çå ¨å¾®åãå 为ï¼
du=(∂u/∂x)dx+(∂u/∂y)dy=y[(3x²-x]e^(1/x)dx+x³e^(1/x)]dy=(2)å¼å·¦è¾¹ã
æ éå½æ° yx³e^(1/x)=Cå°±æ¯åæ¹ç¨çé解ã
(äº)æ±å¾®åæ¹ç¨y''+2y'+y=e^(-x)çé解ã
解ï¼é½æ¬¡æ¹ç¨y''+2y'+y=0çç¹å¾æ¹ç¨ä¸ºr²+2r+1=(r+1)²=0ï¼æéæ ¹r₁=r₂=-1ï¼
æ 该é½æ¬¡æ¹ç¨çé解为y=(C₁+C₂x)e^(-x)
ä¸é¢åæ±ä¸ä¸ªç¹è§£y*ï¼ç¨å¾ å®ç³»æ°æ³ï¼è®¾ç¹è§£ä¸ºy*=ax²e^(-x)
(y*)'=2axe^(-x)-ax²e^(-x)=(2ax-ax²)e^(-x)
(y*)''=(2a-2ax)e^(-x)-(2ax-ax²)e^(-x)=(2a-4ax+ax²)e^(-x)
ä»£å ¥åå¼å¾(2a-4ax+ax²)e^(-x)+2(2ax-ax²)e^(-x)+ax²e^(-x)=e^(-x)
æ¶å»e^(-x)å¾(2a-4ax+ax²)+4ax-2ax²+ax²=1
å³æ2a=1ï¼æ å¾a=1/2ï¼å³æç¹è§£ä¸ºy*=(1/2)x²e^(-x)
æ åæ¹ç¨çé解为ï¼y=(C₁+C₂x)e^(-x)+(1/2)x²e^(-x)=[C₁+C₂x+(1/2)x²]e^(-x)