關於一些工程數學的題目

1.y'x^3 + y(3*x^2) = secx^2 xMdx+Ndy=0 => 正和條件: My=NxM=y(3x^2)-sec^2x => My=3x^2-2sec^2x*tanxN=x^3 => Nx=3x^2=/=My => 非正和 2.y^2 cosx + ( 2ysinx) y' = 2x+1M=y^2*cosx-2x-1 => My=2ycosxN=2ysinx => Nx=2ycosx=My 正和方程式則原式為:0=(y^2*cosx)dx+(2y*sinx)dy-(-2x+1)dx=d(y^2*sinx)-(2x+1)dxc=∫d(y^2*sinx)-∫(2x+1)dx=y^2*sinx-(x^2+x)y^2=(x^2+x+c)/sinx......ans 3.(sin y+ycos x)dx + (sin x + xcos y )dy = 0M=siny+ycosx => My=cosy+cosxN=sinx+x*cosy => Nx=cosx+cosy=My正和則原式為:0=(y*cosx*dx+sinx*dy)+(siny*dx+xcosy*dy)=d(y*sinx)+d(x*siny)c=∫d(y*sinx)+∫d(x*siny)=y*sinx+x*siny.....ans

2014-07-12 10:09:07 補充：
4.(xy^2 - sin x )dx + y(x^2 - 1 )dy = 0

M=xy^2 - sin x => My=2xy-cosx

N=y(x^2 - 1 ) => Nx=2xy=\=My

非正和方程式

2014-07-12 10:10:34 補充：
4.(xy^2 - sin x )dx + y(x^2 - 1 )dy = 0

M=xy^2 - sin x => My=2xy-cosx

N=y(x^2 - 1 ) => Nx=2xy=\=My

非正和方程式

2014-07-12 10:14:34 補充：
5.x^2(6y+1) + (2x^3+3y^2)y' = 0 , y(0)=1

M=(6y+1)x^2 => My=6x^2

N=(2x^3+3y^2) => Nx=6x^2

正和方程式

0=(6yx^2*dx+2x^3*dy)+(x^2*dx+3y^2*dy)

=2[y*d(x^3)+x^3*dy]+d(x^3)/3+d(y^3)

c=2∫d(y*x^3)+∫d(x^3)/3+∫d(y^3)

=2yx^3+x^3/3+y^3......ans

2014-07-12 10:16:16 補充：
代入邊界條件: x=0, y=1

c=0+0+1=1

=> 1=2yx^3+x^3/3+y^3......ans

2014-07-12 10:17:55 補充：
6.1/y + ( -x/y +2y )y' = 0 , y(0)=2

M=1/y => My=-1/y^2

N=-x/y+2y => Nx=-1/y

非正和方程式

2014-07-12 10:26:28 補充：
7.(e^x+y) + ( 2+x+ye^y)y' = 0 , y(0) = 1

M=e^x+y => My=1

N=2+x+ye^y => Nx=1

正和方程式

0=e^x*dx+(ydx+xdy)+(2+y*e^y)dy

c=∫e^x*dx+∫d(xy)+∫(2+y*e^y)dy

=e^x+xy+2y+∫y*d(e^y)

=e^x+xy+2y+{y*e^y-∫e^y*dy}......部份積分

=e^x+xy+2y+y*e^y-e^y

=1+0+2+e-e......代入x=0,y=1

=3

=> 3=e^x+xy+2y+y*e^y-e^y.......ans

2014-07-12 10:31:33 補充：
7.(e^x+y) + ( 2+x+ye^y)y' = 0 , y(0) = 1

M=e^x+y => My=1

N=2+x+ye^y => Nx=1

正和方程式

0=e^x*dx+(ydx+xdy)+(2+y*e^y)dy

c=∫e^x*dx+∫d(xy)+∫(2+y*e^y)dy

=e^x+xy+2y+∫y*d(e^y)

=e^x+xy+2y+{y*e^y-∫e^y*dy}......部份積分

=e^x+xy+2y+y*e^y-e^y

=1+0+2+e-e......代入x=0,y=1

=3

=> 3=e^x+xy+2y+y*e^y-e^y.......ans

2014-07-12 10:34:34 補充：
8.(e^y - 2xy)dx + (xe^y - x^2 )dy = 0 , y(1) = 0

M=e^y - 2xy => My=e^y-2x

N=xe^y - x^2 => Nx=e^y-2x

正和方程式

0=(e^ydx+xe^ydy)-(2xydx+x^2dy)

=d(xe^y)-d(yx^2)

c=∫d(xe^y)-∫d(yx^2)

=xe^y-yx^2

=1-0......x=1,y=0

=1

=> xe^y-yx^2=1......ans

>這家不錯 lｖ3３3。ｃC買幾次啦真的一樣
叩刡兽伉

請核對一下第六題題目。
第一題是 y' x^3 + y(3x^2) = (sec x)^2 嗎？