工程數學求幫忙解答!!謝謝!!?

(1) 0 = (x^2+y^2)dx - 2xydy

= x^2*dx - [x*d(y^2) - y^2*dx]

= dx - [x*d(y^2) - y^2*dx]/x^2


-c = ∫dx - ∫d(y^2/x) = x - y^2/x

y^2/x = c + x

y^2 = x(c+x)

y(x) = +- √[x(x+c)]




(2) y" - 2y' + y = 6x^2*exp(-x)

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

D = 1, 1


yh(x) = (a+bx)*e^x


yp(x) = 6x^2*exp(-x)/(D-1)^2

= e^x∫∫6x^2*exp(-2x)dx*dx

= -6*e^x∫exp(-2x)*(x^2/2 + x/2 + 1/4)dx.....Note

= -6e^x{∫exp(-2x)x^2*dx/2 + ∫exp(-2x)xdx/2 + ∫exp(-2x)dx/4}

= 6e^x*exp(-2x){(x^2/4+x/4+1/8) + (x/4+1/8) + 1/8}

= 6*exp(-x)*(x^2/4 + x/2 + 3/8)

= 3*exp(-x)*(x^2/2 + x + 3/4)


y(x) = (a+bx)*e^x + 3*exp(-x)*(x^2/2 + x + 3/4)


Note: a = -2

∫x*e^(ax)dx = e^(ax)*(x/a-1/a^2) = -exp(-2x)*(x/2 + 1/4)

∫x^2*e(ax)dx =e^(ax)*(x^2/a-2x/a^2+2/a^3)

= -exp(-2x)*(x^2/2 + x/2 + 1/4)


(3) L^-1{2πs/(s^2+π^2)^2} = ?

L{t*f(t)} = -F(s)'

L{sin(πt)} = 1/(s^2+π^2) = F(s)

F(s)' = -2s/(s^2+π^2)^2


L{t*sin(πt)} = 2s/(s^2+π^2)^2

L{πt*sin(πt)} = 2πs/(s^2+π^2)^2


Answer: L^-1{2πs/(s^2+π^2)^2} = πt*sin(πt)




(4) x1 = (0 1 1 1), x2 = (1 0 1 1)


cosQ = x1.x2/|x1*|x2|

= (0 + 0 + 1 + 1)/√(1+1+1)*√(1+1+1)

= 2/√3*√3

= 2/3


Q = acos(2/3)

= 0.666667

= 48.2 degrees