wskl
幼苗
共回答了13个问题采纳率:92.3% 举报
讨论a
(1)a=0
y'=b
y=bt+C
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(2)a不等于0
y'=b-ay^4
dy/[(b/a)-y^4]=adt
讨论b
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(i) b=0
y'/(-y^4)=adt
y^(-3)/3=at+C
y=(3at+C)^(-1/3)
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(ii) b0
dy/((-b/a)+y^4)=-adt
令z=(-b/a)^(1/4)y
dz/(1+z^4)=-a(-b/a)^(-5/4)dt
(1+z^4)=(z^2+1)^2-2z^2=(z^2-根号2 z+1)(z^2+根号2 z +1)
1/(1+z^4)
=(A+Bz)/(z^2-根号2 z+1)+(C+Dz)/(z^2+根号2 z +1)
(A+Bz)/(z^2+根号2 z +1)+(C+Dz)/(z^2-根号2 z+1)=1
A+C=1
B+D=0
A+根号2 B+C-根号2 D=0
根号2 A+B- 根号2 C+D=0
A=C=1/2,B=-根号2/4,D=根号2/4
积分dz/(1+z^4)
=积分(1/2-(根号2/4)z)/(z^2+根号2 z +1)+(1/2+(根号2/4)z)/(z^2-根号2 z+1)
=-(根号2/8)积分 (2z+根号2)dz/(z^2+根号2 z +1)
+(3/4)积分dz/[(z+根号2/2)^2+1/2]
+(根号2/8)积分 (2z-根号2)dz/(z^2-根号2 z +1)
+(3/4)积分dz/[(z-根号2/2)^2+1/2]
=-(根号2/8) ln|z^2+根号2 z +1|+(3/4)*[1/根号(1/2)]*arctan[(z+根号2/2)/(根号(1/2))]
+(根号2/8) ln|z^2-根号2 z +1|+(3/4)*[1/根号(1/2)]*arctan[(z-根号2/2)/(根号(1/2))]
=-a(-b/a)^(-5/4)t+C
把z=(-b/a)^(1/4)y代入即可
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(iii)b>0
b/a>0
dy/((b/a)-y^4)=adt
令z=(b/a)^(1/4)y
dz/(1-z^4)=a(b/a)^(-5/4)dt
1/(1-z^4)=(1/2)[1/(1-z^2)+1/(1+z^2)]
=(1/4)[1/(1-z)+1/(1+z)]+(1/2)[1/(1+z^2)]
积分得到
(1/4)ln|(1+z)/(1-z)|+(1/2)arctan z=a(b/a)^(-5/4)t+C
把z=(b/a)^(1/4)y代入即可
1年前
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