Anonymous
Anonymous asked in 教育與參考考試 · 1 decade ago

請問ODE的四題正合相關問題?

1.y'+(x+1)y=e^(x^2)y^3,y(0)=0.5

2.y'sin2y+xcos2y=2x

3.2yy'+y^2sinx=sinx,y(0)=√2

4.y'+x^2y=[e^(-x^3)sinhx]/(3y^2)

就這四題請高手們幫我看看好嗎?

Update:

using a method of this section or separating variables,find the general solution.if an initial condition is given,find also the particular solution and sketch or graph it.

2 Answers

Rating
  • 龍昊
    Lv 7
    1 decade ago
    Favorite Answer

    確定這些題目都是正合( exact )的題目嗎?

    2006-10-22 00:35:31 補充:

    using a method of this section,意思是使用這一節的方法,能不能跟我講是哪一節、名稱是什麼?因為有的題目用紙筆算不出來,是不是要用數值分析、數值方法?

    2006-10-22 12:25:26 補充:

    1. y' + ( x + 1 ) y = exp( x2 ) y3 , y(0) = 0.5sol:  原式為 Bernoulli's ode,移項得:y - 3 + ( x + 1 ) y - 2 = exp( x2 )  令 u = y - 2 → u' = - 2y - 3 y'  → y - 3 y' = - ( u'/2 )  → - ( u'/2 ) + ( x + 1 ) u = exp( x2 )  → u' - 2( x + 1 ) u = - 2 exp( x2 ) ~ 一階線性 o.d.e.  積分因子:I(x) = e∫- 2( x + 1 )dx = exp( - x2 - 2x )  u = exp( x2 + 2x )[ - 2∫exp( - x2 - 2x ) exp( x2 )dx + c ]     = exp( x2 + 2x )[ - 2∫exp( - 2x )dx + c ]     = exp( x2 + 2x )[ exp( - 2x ) + c ]     = exp( x2 ) + c exp( x2 + 2x )  → u = exp( x2 ) + c exp( x2 + 2x )  → y - 2 = exp( x2 ) + c exp( x2 + 2x )  y(0) = 0.5 → x = 0 , y = 0.5  → 4 = e0 + ce0 → 4 = 1 + c  → c = 3  → y - 2 = exp( x2 ) + 3 exp( x2 + 2x ) #*2. y' sin 2y + x cos 2y = 2xsol:  ( dy/dx )( sin 2y ) + x cos 2y = 2x  → ( dy/dx )( sin 2y ) = x( 2 - cos 2y )  → [ ( sin 2y )/( 2 - cos 2y ) ]dy = xdx  →∫[ ( sin 2y )/( 2 - cos 2y ) ]dy =∫xdx + c1  → ( 1/2 )∫[ 1/( 2 - cos 2y ) ]d( 2 - cos 2y ) =∫xdx + c1  → ( 1/2 ) ln│2 - cos 2y│= ( x2/2 ) + c1  → ln│2 - cos 2y│= x2 + 2c1  令 2c1 = c  → ln│2 - cos 2y│= x2 + c #*3. 2yy' + y2sin x = sin x , y(0) = √2sol:  原式為 Bernoulli's o.d.e.  令 u = y2 → u' = 2yy'  → u' + ( sin x ) u = sin x ~ 一階線性 o.d.e.  積分因子:I(x) = e∫sin xdx = e - cos x  u = e cos x (∫e - cos x sin x dx + c )     = e cos x [ -∫e - cos x d( cos x ) + c ]     = e cos x [ e - cos x + c ]     = 1 + ce cos x  → u = 1 + ce cos x  → y2 = 1 + ce cos x  y(0) = √2 → x = 0 , y = √2  → 2 = 1 + ce1  → c = ( 2 - 1 )/e1 = e - 1  → y2 = 1 + e - 1e cos x #*4. y' + x2y = [ exp( - x3 )( sinh x ) ]/3y2sol:  原式為 Bernoulli's o.d.e.,移項得:y - 2 y' + x2y - 1 = [ exp( - x3 )( sinh x ) ]/3  令 u = y - 1 → u' = - y - 2 y'  → - u' + x2u = exp( - x3 )( sinh x )/3  → u' - x2u = - exp( - x3 )( sinh x )/3 ~ 一階線性 o.d.e.  積分因子:I(x) = exp(∫- x2dx ) = exp( - x3/3 )  u = exp( x3/3 ) [ - ( 1/3 )∫exp( - x3/3 )exp( - x3 )( sinh x )dx + c ]     = exp( x3/3 ) [ - ( 1/3 )∫exp( - 4x3/3 )( sinh x )dx + c ]  卡在這個積分∫exp( - 4x3/3 )( sinh x )dx 算不出來,抱歉囉!*  希望以上回答能幫助您。

    Source(s): 自己
  • Anonymous
    1 decade ago

    using a method of this section or separating variables,find the general solution.if an initial condition is given,find also the particular solution and sketch or graph it.

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