✔ 最佳答案
1.0=(1-x^2)y"-4xy'-2y=[(1-x^2)y']'-2(xy'+y)=[(1-x^2)y']'-2(xy)'=[(1-x^2)y'-2(xy)]'=(1-x^2)y'-2(xy)=(1-x^2)dy-2xydx=dy/y-2xdx/(1-x^2)ln(c)=ln(y)+d(1-x^2)/(1-x^2)=ln(y)+ln(1-x^2)c=y(1-x^2)
y(x)=c/(1-x^2)......ans
2.y"+λ*y=0, y(0)=0 & y'(L)=0(2A)λ>0 => λ=a^20=D^2+a^2 => D=+-ajy(t)=c1*cos(at)+c2*sin(at)y(0)=c1=0 => y(x)=c2*sin(at)y'(x)=a*c2*cos(at)y'(L)=a*c2*cos(aL)=0=> cos(aL)=0 => a*L=(2n-1)*pi/2; n=自然數L=(2n-1)*pi/2a
(2B)λ<0 => λ=-a^20=D^2-a^2 => D=+-ay(t)=c1*e^t+c2*e(-t) => y(0)=c1+c2=0y'(t)=c1*e^t-c2*e^(-t) => y'(L)=c1*e^L-c2*e^(-L)=0=> c1=c2=0所以λ<0不存在(2C)λ=0y"=0 => y(x)=c1*x+c2
3.y"+8y'+(a+16)y=0, y(0)=0 & y(π)=0(3A)λ>0 => λ=a^20=D^2+8D+(a^2+16)D=-4+-ajy(t)=e^(-4t)[c1*cos(at)+c2*sin(at)]y(0)=c1=0=> y(t)=c2*e^(-4t)*sin(at)y'(t)=c2{-4e^(-4t)*sin(at)+a*e^(-4t)*cos(at)}=> y'(pi)=c2{-4e^(-4pi)*sin(a*pi)+a*e^(-4pi)*cos(a*pi)}=0=> 0=-4*sin(a*pi)+a**cos(a*pi)=> tan(a*pi)=a/4 => a=0 => λ=0(3B)λ<0 => 同樣的道理,不存在(3C)λ=0 => 參考3A
2014-04-09 20:12:28 補充：
第1題漏掉積分常數修正如下:
c1=(1-x^2)y'-2(xy)
c1/(1-x^2)=y'-2xy/(1-x^2)
積分因子:
ln(F)=-2∫xdx/(1-x^2)
=∫d(1-x^2)/(1-x^2)
=ln(1-x^2)
=> F=(1-x^2)
代入原式:
[y*(1-x^2)]'=[c1/(1-x^2)]*(1-x^2)=c1
y(1-x^2)=∫c1*dx+c2=c1*x+c2
y(x)=(c1*x+c2)/(1-x^2)......ans
