請問ODE的四題正合相關問題?

Question

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?

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.

Answer 1

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。

Answer 2

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.