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1. -πsinπx sinh ydx+ cosπx cosh ydy= 0
2. (e^y-ye^x)dx+ (xe^y-e^x)dy= 0
3. e^-2θdr - 2re^-2θdθ= 0
4. (cosωx+ ω sinωx )dx+ e^x dy= 0 ,y(0)=1
5. (cosxy + x/y )dx+ (1+(x/y)cosxy)dy = 0
6. ( x^4+y^2)dx -x/y dy = 0 y(2)=1

請大大ㄇ幫助我解題謝謝

2006-10-01 19:17:59 · 2 個解答 · 發問者 光華 1 in 教育與參考 考試

2 個解答

1. - π sin πx sinh y dx + cos πx cosh y dy = 0sol:  由題目移項得:cos πx cosh y dy = π sin πx sinh y dx  → ( cosh y/sinh y )dy = π ( sin πx/cos πx )dx  →∫coth y dy = π∫tan πx dx + c1  → ln│sinh y│= - ln│cos πx│+ c1  等號兩邊同取指數 → sinh y = ( ec1/cos πx )  令 ec1 = c  → ( sinh y )( cos πx ) = c #* 2. ( ey - yex )dx + ( xey - ex )dy = 0sol:  令 M( x , y ) = ey - yex    N( x , y ) = xey - ex  ( ∂M/∂y ) = ( ∂N/∂x ) = ey - ex  → 正合,必有一解,其型式為:F( x , y ) = C  F =∫Mdx + c(y) =∫( ey - yex )dx + c(y)     = xey - yex + c(y)  ( ∂F/∂y ) = N → xey - ex + c'(y) = xey - ex  → c'(y) = 0 → c(y) = 0  → xey - ex = C # *3. e - 2θdr - 2re - 2θdθ = 0sol:  由題目移項得:e - 2θdr = 2re - 2θdθ  →∫( 1/r )dr =∫2dθ + c1  → ln│r│= 2θ + c1  等號兩邊同取指數 → r = e2θec1  令 ec1 = c  → r = ce2θ # *4. ( cos ωx + ω sin ωx )dx + exdy = 0 , y(0) = 1sol:  原是移項得:exdy = - ( cos ωx + ω sin ωx )dx  →∫dy = -∫e - x( cos ωx + ω sin ωx )dx + c  → y = [ e - x/( 1 + ω2 ) ]( cos ωx + 2ω sin ωx + ω2cos ωx ) + c  y(0) = 1 → x = 0 , y = 1  → 1 = [ 1/( 1 + ω2 ) ]( 1 + 0 + ω2 ) + c  → c = 0  → y = [ e - x/( 1 + ω2 ) ]( cos ωx + 2ω sin ωx + ω2cos ωx ) # *5. [ cos xy + ( x/y ) ]dx + [ 1 + ( x/y )cos xy ]dy = 0sol:  這題我算不出來 >_<* 6. ( x4 + y2 )dx - xydy = 0 , y(2) = 1sol:  原題目同除 x2 → [ x2 + ( y/x )2 ]dx - ( y/x )dy = 0  → ( y/x )( dy/dx ) = x2 + ( y/x )2  → 令 ( y/x ) = u → y = xu          → ( dy/dx ) = u + x( du/dx )  → u [ u + x( du/dx ) ] = x2 + u2  → x( du/dx ) = ( x2/u )  →∫udu =∫xdx + c1  → ( u2/2 ) = ( x2/2 ) + c1  → u2 = x2 + 2c1  令 2c1 = c  → ( y/x )2 = x2 + c  y(2) = 1 → x = 2 , y = 1  → ( 1/4 ) = 4 + c   → c = ( - 15/4 )   → ( y/x )2 = x2 - ( 15/4 ) #*  希望以上回答內容能幫助您。

2006-10-02 19:36:18 · answer #1 · answered by 龍昊 7 · 0 0

5. (cosxy + x/y )dx + [1 +(x/y)cosxy]dy = 0
sol:
等號兩邊同乘y

(ycosxy + x)dx + (y + xcosxy)dy=0

cosxy(ydx + xdy) + xdx + ydy=0

cosxyd(xy) + xdx + ydy=0

sinxy + 1/2x^2 + 1/2y^2=1/2C

2sinxy + x^2 + y^2=C

2006-10-04 13:21:34 · answer #2 · answered by Ian 3 · 0 0

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