設函數y=f(x)滿足方程式x(dy/dx)=x^2+3y ,(x>0)且滿足條件
y(2)=4 ,求此函數為何??
答案y=-x^2+x^3有沒有人會算
有關函數的微分問題(有一點偏微)
2009-06-13 2:21 am
回答 (3)
2009-06-13 2:54 am
✔ 最佳答案
x(dy/dx)=x2+3y xdy=x2dx +3ydx ⇒xdy-3ydx=x2dx (在等號兩邊同乘x2)
x3dy-3yx2dx=x4dx , 因此 x3dy-ydx3=x4dx (在等號兩邊同乘x6)
d( y/x3)=x-2dx , 積分得: y/x3 =(-1/x)+c
(因為 d( y/x3)=(x3dy-ydx3)/x6 )
所以 y=-x2 +cx3
而y(2)=4 , 4=-4+8c, 8=8c,c=1
所以 y=-x2 +x3
2009-06-13 3:06 am
x(dy/dx)=x^2+3y
(dy/dx)-(3/x)y=x
P(x)=-3/x,Q(x)=x.∫P(x) dx =-3lnx
Integration factor is x^(-3)
So y=[∫(x^-3)(x) dx + C ]x^3 =[-x^(-1) + C]x^3
Substitute y(2)=4
[(-1/2)+C]8=4=>C= 1
So y=-x^2+x^3
(dy/dx)-(3/x)y=x
P(x)=-3/x,Q(x)=x.∫P(x) dx =-3lnx
Integration factor is x^(-3)
So y=[∫(x^-3)(x) dx + C ]x^3 =[-x^(-1) + C]x^3
Substitute y(2)=4
[(-1/2)+C]8=4=>C= 1
So y=-x^2+x^3
2009-06-13 2:49 am
x(dy/dx)=x^2+3y
=>dy/dx-(3/x)*y=x
積分因子I(x)=e^[-3*∫(1/x)dx]=e^[-3ln(x)]=1/(x^3)
I(x)*[dy/dx-(3/x)*y]=I(x)*x
=>[1/(x^3)]*(dy/dx)-3/(x^4)=1/(x^2)
=>d[x^(-3)*y]/dx=1/(x^2)
=>x^(-3)*y=-1/x+c
=>y=-x^2+cx^3
將x=2代入上式得c=1
故y=-x^2+x^3
=>dy/dx-(3/x)*y=x
積分因子I(x)=e^[-3*∫(1/x)dx]=e^[-3ln(x)]=1/(x^3)
I(x)*[dy/dx-(3/x)*y]=I(x)*x
=>[1/(x^3)]*(dy/dx)-3/(x^4)=1/(x^2)
=>d[x^(-3)*y]/dx=1/(x^2)
=>x^(-3)*y=-1/x+c
=>y=-x^2+cx^3
將x=2代入上式得c=1
故y=-x^2+x^3
收錄日期: 2021-04-26 13:59:37
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