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1.y"'+3y"-4y=0

輔助方程式:

0=D^3+3D^2-4=(D-1)(D+2)^2

D=1,-2,-2

So y(x)=(c1+c2*x)*e^(-2x)+c3*e^x

2.y"-10y'+25y=30x+3

輔助方程式:

0=D^2-10D+25=(D-5)^2

D=5,5

齊次解: yh(x)=(c1+c2*x)*e^(5x)

設特殊解:

yp(x)=a*x^2+bx+c => 25yp=25a*x^2+25bx+25c

yp'=2ax+b => -10yp=-20ax-10b

yp"=2a

Add: 30x+3=25ax^2+(25b-20a)x+(25c-10b+2a)

a=0, 25b-20a=30 => b=6/5

25c-10b+2a=3 => c=3/5

So yp(x)=(2x+1)3/5

通解: y(x)=yh(x)+yp(x)=(c1+c2*x)*e^(5x)(2x+1)3/5

3.IF F=(x^2*y^3-z^4)i+(4x^5*y^2*z)j-(y^4*z^6)k, find

(a)curl(F)

=|∂/∂y ..........∂/∂z .....∂/∂x ..........∂/∂y........|
.|(4x^5*y^2*z) -(y^4*z^6) (x^2*y^3-z^4) (4x^5*y^2*z)|

=(-4y^3*z^6-4x^5*y^2; -4z^3+0; 20x^4*y^2*z-3x^2*y^2)

=-4y^2(y*z^6+x^5)i-4z^3j+x^2*y^2(20x^2+3)k

(b)div(F)

=∂Fx/∂x+∂Fy/∂y+∂Fz/∂z

=2xy^3+8x^5*yz-6y^4*z^5

=2y(xy^2+4x^5*z-3y^3*z^5)

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