求工程數學解答

1.(x^2+3y^2)dx-2xydy=0齊次方程式可以令y=ux => y'=u+u'x代入上式裡面:0=(x^2+3u^2x^2)-2xux(u'x+u)=(1+3u^2)-2u(u'x+u)=(1+3u^2)-2uu'-2u^2=(1+u^2)-2udu/dx=(1+u^2)dx-2ududx=2udu/(u^2+1)∫dx=∫d(u^2+1)/(u^2+1)x+c1=ln(u^2+1)=ln(y^2/x^2+1)=ln[(x^2+y^2)/x^2]e^(x+c1)=(x^2+y^2)/x^2x^2+y^2=c*x^2*e^x.....c=e^c1y^2=x^2(c*e^x-1)y(x)=+-x√(ce^x-1).....ans

2.(3x+y-1)dx-(6x+2y+3)dy=0使用取代法.令u=3x+y => y'=u'-3代入上式裡面:0=(u-1)-(2u+3)(u'-3)=u-1-(2u+3)u'+3(2u-3)=(7u-10)-(2u+3)du/dxdx=(2u+3)du/(7u-10)∫dx=(2/7)∫du+41∫du/7(7u-10)x+c1=2u/7+(41/49)*ln(7u-10)=2(3x+y)/7+(41/49)*ln[7(3x+y)-10]7(x+c1)=2(3x+y)+(41/7)*ln[7(3x+y)-10]7(x-2y+7c1)/41=ln[7(3x+y)-10].....c=7c17(3x+y)-10=e^[7(x-2y+c)/41]......ans

1.(x^2+3y^2)dx-2xydy=0齊次方程式可以令y=ux => dy=udx+xdu代入上式裡面:0=(x^2+3(ux)^2)dx-2xux(udx+xdu)=(1+3u^2-2u^2)dx-2(ux)du

[2u/(1+u^2)]du=(1/x)dx ===> ln(u^2+1)=k+ln(x)=ln(cx)

u^2+1=cx ===> y^2=(cx-1)

2.可令 3x+y=u (du=3dx+dy)

原式變成 (u-1)dx-(2u+3)(du-3dx)=0 ====> (7u+8)dx=(2u+3)du

dx=[(2u+3)/(7u+8)]du=[2/7+(5/49)/(u+8/7)]du ====> 49x=14u+5ln(u+8/7)

49x=14(3x+y)+5ln(3x+y+8/7) ====> (7/5)x=(14/5)y+ln(3x+y+8/7)

解到這裡就可以了(已經是顯函數解)


2013-11-08 02:06:50 補充：
第1題
最後帶錯了 應該是 y^2=(cx-1)(x^2) 才對

驗算時才發現不對