求微分問題和積分問題的解.(3)

(1) y=(10-5x^3)/x => xy"+2y'+30x=0 => 不成立0=x*dy'/dx+2y'+30x=xu'+2u+30x <u=y'>=u'+2u/x+30積分因子 => Ln(F)=∫2dx/x=Ln(x^2)F=x^2 => u=(-30/x^2)∫x^2dx+cu=y'=(-30/x^2)x^3/3+c=-10x+c => y"=-10.....ay=∫(-10x+c)dx+c1=-5x^2+cx+c1與y=(10-5x^3)/x顯然不同Eq.a 代入原微分方程式: 0=-10x+2(-10x+c)+30x=2c => c=0所y應該改為: y=-5x^2+c1
(2a) y=x√(6+3x^2)y'=√(6+3x^2)+(x/2)*(6x)/√(6+3x^2)=(6+3x^2+3x^3)/√(6+3x^2)=6(1+x^2)/√(6+3x^2)(2b)∫(1+x^2)dx/√(6+3x^2) <x=1~5>=∫y'dx/6=x√(6+3x^2)/6=5√(6+3*25)/6-√(6+3)/6=5√81/6-3/6=(5*9-3)/6=42/6=7..........ans

(a) y = (10 - 5x^3)/x

dy/dx

= [x(-15x^2) - (10 - 5x^3)]/x^2

= (-10x^3 - 10)/x^2

d^2y/dx^2

= [x^2(-30x^2) - (-10x^3 - 10)(2x)]/x^4

= (-10x^4 + 20x)/x^4

因此﹐x(d^2y/dx^2) + 2(dy/dx) + 30x

= (-10x^4 + 20x)/x^3 + 2(-10x^3 - 10)/x^2 + 30x

= -10x + 20/x^2 - 20x - 20/x^2 + 30x

= 0

(b)(i) dy/dx

= 3x^2/√(6 + 3x^2) + √(6 + 3x^2)

= (3x^2 + 6 + 3x^2)/√(6 + 3x^2)

= (6x^2 + 6)/√(6 + 3x^2)

(ii) ∫ (1->5) (1 + x^2)/√(6 + 3x^2) dx

= (1/6) ∫ (1->5) (6 + 6x^2)/√(6 + 3x^2) dx

= (1/6) ∫ (1->5) dy/dx dx [當中y = x√(6 + 3x^2)]

= (1/6) ∫ (3->45) dy

= (1/6)(45 - 3)

= 7