幫幫忙~ 工程數學高手請進

Question

1.試解微分方程式(X(2次) 1)dx(分之)dy 4xy=X
2.試解含初期值之微分方程式{X dx(分之)dy-2y=2x(4次方)
{y(2)=8
3.試利用合併法求解下列微分方程式
A.[X(x<2次方> y<2次方>)2次-y] dx [(x<2次方> y<2次方> )2次y x] dy=0
B.Y(x y)dx x (x y)dy dx dy=0
4.利用柏努力方程式求解A.dx(分之)dy y=(xy)2次方
B.X dx(分之)dy=y xy3次方(HlinX)
5.試解(x<2次方>-1)dy (y<2次方>-2xy 1)dx=0
6.有一RL電路.外加電壓為5伏特.其電阻50歐姆.電感為1H T=0時,(符號像舒寫體的T) I=0,球再時間T十的電流工?

Answer 1

1. (x2+1)(dy/dx)+4xy=x  => y'+[4x/(x2+1)]y=x/(x2+1)   積分因子I=e∫[4x/(x^2+1)]dx=e2*ln(x^2+1)=(x2+1)2   乘到原式 (x2+1)2y'+[4x(x2+1)]y=x/(x2+1)   => [(x2+1)2y]'=x/(x2+1)   => (x2+1)2y=∫[x/(x2+1)]dx+C             =(1/2)ln(x2+1)+C   => y=(1/2)(x2+1)-2ln(x2+1)+C(x2+1)-22. x(dy/dx)-2y=2x4, y(2)=8   y'-(2/x)=2x3   積分因子I=e-∫(2/x)dx=e-2*lnx=x-2   乘到原式 x-2y'-(2/x3)y=2x   => (x-2y)'=2x   => x-2y=∫2xdx+C          =x2+C   => y=x4+Cx2  y(2)=8 => 8=24+C*22 =>C=-2   y=x4-2x23.A. [x(x2+y2)2-y]dx+[(x2+y2)2y+x]dy=0   x(x2+y2)2dx-ydx+(x2+y2)2ydy+xdy=0   (x2+y2)2 (xdx+ydy)+(xdy-ydx)=0   (1/2)(x2+y2)2d(x2+y2)+x2d(y/x)=0 B. y(x+y)dx+x(x+y)dy+dx+dy=0    (x+y)(ydx+xdy)+d(x+y)=0    (x+y)d(xy)+d(x+y)=0    d(xy)+[d(x+y)]/(x+y)=0    ∫d(xy)+∫[d(x+y)]/(x+y)=C    xy+ln|x+y|=C4.A. (dy/dx)+y=(xy)2    y'+y=x2y2   令 u=y1-2=y-1 => u'=-y-2y' => y'=-y2u'    代入原式    -y2u'+y=x2y2 => u'-y-1=-x2   => u'-u=-x2   積分因子 I=e-∫dx=e-x   乘到原式    e-xu'-e-xu=-x2e-x    (e-xu)'=x2e-x    e-xu=∫x2e-x+C        =-x2e-x-2xe-x-2e-x+C    u=-x2-2x-2+Cex    y-1=-x2-2x-2+CexB. x(dy/dx)=y+xy3   y'-(1/x)y=y3   令 u=y1-3=y-2 => u'=-2y-3y' => y'=-(1/2)y3u'   代入原式    -(1/2)y3u'-(1/x)y=y3     u'+(2/x)y-2=-2     u'+(2/x)u=-2    積分因子 I=e∫(2/x)dx=e2lnx=x2    乘到原式 x2u'+(2x)u=-2x2     (x2u)'=-2x2     x2u=-∫2x2dx+C=-(2/3)x3+C     u=-(2/3)x+Cx-25. (x2-1)dy+(y2-2xy+1)dx=0   (x2-1)y'+y2-2xy+1=0   有一解 y1=x   令 y=y1+u-1=x+u-1     y'=1-u-2u'    代入原式 (x2-1)(1-2u-2u')+(x+u-1)2-2x(x+u-1)+1=0     => x2-1-(x2-1)u-2u'+x2+2xu-1+u-2-2x2-2xu-1+1=0     => -(x2-1)u-2u'+u-2=0     => -(x2-1)u'+1=0     => (x2-1)(du/dx)=1     => du=dx/(x2-1)=(1/2)[-1/(x+1)+1/(x-1)]dx     => ∫du=(1/2)∫[-1/(x+1)+1/(x-1)]dx+lnC     => u=(1/2)(-ln|x+1|+ln|x-1|)+lnC         =(1/2)ln|(x-1)/(x+1)|+lnC         =ln[C|(x-1)/(x+1)|1/2]     => y=x+u-1=x+1/ln[C|(x-1)/(x+1)|1/2]6. L(di/dt)+Ri=v  => (di/dt)+50i=5  => di/dt=5-50i  => di/(5-50i)=dt  => ∫di/(5-50i)=∫dt+C'  => -(1/50)ln|5-50i|=t+C'  => ln|5-50i|=-50t-50C'  => 5-50i=e-50te-50C'=C"e-50t  (C"=e-50C')  => i=(5-C"e-50t)/50=0.1+Ce-50t  (C=-C"/50)