# 微分方程一題

y''-4xy'+4x^2y=xe^x^2

Update:

(D-2x+√2i)(D-2x-√2i)y是怎麼分的

Update 2:

Update 3:

to 麻辣

Rating
• 7 years ago
• 7 years ago
• 7 years ago

原式 = [ ( D - 2x )^2 +2 ]y

= [ ( D - 2x )^2 - ( √2i )^2 ]y

= ( D - 2x + √2i )( D - 2x - √2i )y

• 7 years ago

to 麻辣:

integrate tan x <> sec^2 x

integrate tan x = - ln (cos x)

• 麻辣
Lv 7
7 years ago

y"-4xy'+4x^2y=xe^x^2 (1) 齊次方程式y"-4xy'+4x^2y=0令y=e^t => y'=e^t*t'y"=e^t(t"+t'^2)代入上式裡面: 0=(t"+t'^2)-4xt'+4x^2

令t'=u+2x => t"=u'+2代入上式裡面: 0=(u'+2)+(u^2+4xu+4x^2)-4x(u+2x)+4x^2=u'+u^2+2=du/(u^2+2)+dx=∫du/(u^2+2)+∫dx+c1=[atan(u/√2)]/√2+x-c1(c1-x)√2=atan(u/√2)tan[(c1-x)√2]=u/√2t'-2x=√2*tan[(c1-x)√2]t=2∫xdx+√2∫tan[(c1-x)√2]dx+ln(c2)=x^2-∫tan[(c1-x)√2]d[(c1-x)√2]+ln(c2)ln(y)=x^2-sec²[(c1-x)√2]+ln(c2)yh(x)=c2*exp{x^2-sec²[(c1-x)√2]}

(2) 特殊解yp=a*x*e^(x^2)yp'=e^(x^2)*(a+2ax^2)yp"=e^(x^2)*(6ax+4ax^3)代入原式裡面,並把e^(x^2)刪除:x=(6ax+4ax^3)-4x(a+2ax^2)+4ax^2=2axa=1/2=> yp(x)=x*e^(x^2)/2

(3) 一般解y(x)=yh+yp=c2*exp{x^2-sec²[(c1-x)√2]}+x*e^(x^2)/2......ans

2014-08-28 07:41:15 補充：

與版主答案有異

但是過程似乎沒有錯誤

有請高人指點

2014-08-28 16:51:46 補充：

感恩自由自在老師的指導

tan積分修改:

t=2∫xdx+√2∫tan[(c1-x)√2]dx+ln(c2)

=x^2-∫tan[(c1-x)√2]d[(c1-x)√2]+ln(c2)

ln(y)=x^2-ln{sec[(c1-x)√2]}+ln(c2)

ln{ysec[(c1-x)√2]/c2}=x^2

y*sec[(c1-x)√2]=exp(c^2*x^2)

yh(x)=c3*cos[√2(c1-x)]*exp(x^2).......c3=exp(c^2)

2014-08-28 16:52:12 補充：

一般解

y(x)=yh+yp

=c3*cos[√2(c1-x)]*exp(x^2)+x*e^(x^2)/2

=exp(x^2){c3*cos[√2(c1-x)]+x}......ans

2014-08-28 16:56:13 補充：

=exp(x^2){c3*cos(√2c1)*cos(√2x)+c3*sin(√2c1)*sin(√2x)}

=exp(x^2){c4*cos(√2x)+c5*sin(√2x)}........ans

=版主答案

2014-08-28 16:58:23 補充：

漏打x/2補充:

y(x)=exp(x^2)*{c4*cos(√2x) + c5*sin(√2x) + x/2}........ans

c4=c3*cos(√2c1)

c5=c3*sin(√2c1)

• 7 years ago

沒有這麼複雜答案是y=e^x^2(c1cos√2x+c2sin√2x+x/2)

2014-08-28 10:48:13 補充：

你很厲害喔

我剛剛還在找他哪裡有錯

排版實在太難看了

• ?
Lv 7
7 years ago