淡江大學工程數學轉學考問題

1.
(2xy^2 - y)dx+(2x - yx^2)dy = 0
(∂/∂y)(2xy^2-y)= 4xy-1
(∂/∂x)(2x-yx^2)= 2-2xy
兩者不相等,故 nonexact
但相差 6xy-3與 2xy^2-y的比值為y的函數,則有積分因子u(y)
設 u(y)*(2xy^2-y) dx+ u(y)*(2x-yx^2) dy=0為exact, 則
(∂/∂y)[u(y)*(2xy^2-y)]= (∂/∂x)[u(y)*(2x-yx^2)
u'(y)*(2xy^2-y)+u(y)*(4xy-1)= u(y)*(2-2xy)
u'(y)*y(2xy-1)= 3(1-2xy)*u(y)
u'(y)/u(y) = -3/y 為separable eq.
兩邊積分,得 ln[u(y)]= -3ln(y)=ln(1/y^3), 故取u(y)= 1/y^3
原ODE改為 (1/y^3)(2xy^2-y)dx+(1/y^3)(2x-yx^2) dx=0
(2x/y - 1/y^2) dx+ (2x/y^3- x^2/y^2) dy=0 為exact eq.
∫ (2x/y- 1/y^2) dx= x^2/y - x/y^2 + C1(y)
∫ (2x/y^3- x^2/y^2) dy= -x/y^2 + x^2/y + C2(x)
兩相比較得原ODE通解為 x^2/y - x/y^2 = C

2. y"+4y=8sin(2x)
(1)先解y"+4y=0
輔助方程式 t^2+4=0, t=2i, -2i
則齊次解為 yh(x)= Acos(2x)+Bsin(2x)
(2)再求特解
設y=x[Acos(2x)+Bsin(2x)]滿足y"+4y=8sin(2x)
(註:因8sin(2x)為齊次解,故多乘以x)
得 (4B-4Ax)cos(2x)+(-4A-4Bx)sin(2x)+ 4x[Acos(2x)+Bsin(2x)]= 8sin(2x)
4B=0, -4A=8, 則A=-2, B=0
得特解yp= -2xcos(2x)
(3)原ODE y"+4y=8sin(2x)通解為 y=yh+yp=Acos(2x)+Bsin(2x)- 2xcos(2x)

3.
(1)先找eigenvalue
det(A-xI)=0, 則 (-1-x)(3-x)=0, x= -1, 3
(2) x= -1時 (A+I)(x,y)^t=(0,0)^t, 即
[0 4][x]=[0]
[0 4][y] [0]
取(x,y)=(1, 0), 得eigenvalue -1之相對 eigenvector為[1, 0]^t
(3) x=3 時, (A-3I)(x,y)^t=(0,0)^t, 即
[-4 4] [x] = [0 ]
[ 0 0] [y] [0 ]
取(x,y)=(1, 1), 得eigenvalue 3之相對 eigenvector為[ 1 1 ]^t