令y?u 则
y?ux
dy=u?xdu
xdx因此：u?xdudx=xu2
du1 dx?u2 u2du?dx 13u3?x?c
u3?3x?x?c
将yx?u带入 （*）中 dx*）
得：- 11 -
y3?3x4?cx3是原方程的
（ dy2y??(x?1)3dxx?1dy2y解：??(x?1)3dxx?12P(x)?,Q(x)?(x?1)3x?17.e?P(x)dx?e?x?1dx2?(x?1)2方程的通解为：P(x)dx?P(x)dx y=e?(?e?Q(x)dx?c)2 =(x+1)(?13*(x+1)dx+c)(x?1)22 =(x+1)(?(x+1)dx+c) (x?1)2 =(x+1)(?c)2 即：2y=c(x+1)2+(x+1)4为方程的通解。2dyy 8. =dxx?y3dxx+y31解：??x?y2dyyy1则P(y)=,Q(y)?y2y e?P(y)dy?e?ydy1?y方程的通解为：P(y)dy?P(y)dy x=e?(?e?Q(y)dy?c)1 =y(?*y2dy?c)yy3 =?cy23y解.即 x= +cy是方程的通解 ，且y=0也是方程的解。 2 - 12 -
dyayx?1??,a为常数dxxxax?1解：（Px)?,Q(x)?xx9.e?P(x)dxdx?x?e?xaP(x)dx?P(x)dxa方程的通解为： y=e?(e?Q(x)dx?c)1x+1 =xa(?adx+c) xx 当 a?0时，方程的通解为 y=x+ln/x/+c 当 a?1时，方程的通解为 y=cx+xln/x/-1 当 a?0，1时，方程的通解为x1 y=cxa+- 1-aa 10.xdy?y?x3dxdy1解：??y?x3dxx1P(x)??,Q(x)?x3x1??dxP(x)dx1e??ex?x方程的通解为： y=e?P(x)dx?P(x)dx(?e?Q(x)dx?c)1 =(?x*x3dx?c)xx3c =?4xx3c方程的通解为： y=?4x
- 13 -
11.dy?xy?x3y3dxdy解：??xy?x3y3dx两边除以y3dy?23??xy?xy3dxdy-2??2(?xy?2?x3)dx令y?2?zdz??2(?xz?x3)dxP(x)?2x,Q(x)??2x3e?p?x?dx?e?2xdx?ex2方程的通解为： z= e?2p?x??p?x?dx(?e?dxQ(x)dx?c)2 =ex(?e?x(?2x3)dx?c) =x2?cex?1故方程的通解为：y2(x2?cex?1)?1,且y?0也是方程的解。12.(ylnx?2)ydx?xdydylnx22y解：?y?dxxx2 两边除以ydylnx2y?1??y2dxxxdy?1lnx2y?1??dxxx 令y?1?zdz2lnx?z?dxxx2lnxP(x)?,Q(x)??xx方程的通解为：z?e?P(x)dx?P(x)dx(?e?Q(x)dx?c)22c2lnx1x??42422dx??dxlnx1lnx?xz?e(?ex(?)dx?c)?x2(?2(?)dx?c)xxxclnx1?x2??424clnx1方程的通解为:y(x2??)?1,且y=0也是解。424
- 14 -
13
2xydy?(2y2?x)dxdy2y2?xy1???dx2xyx2y
这是n=-1时的伯努利方程。 两边同除以1，
ydyy21y?? dxx2令y2?z
dzdy?2y dxdxdz2y22z??1??1 dxxxP(x)=2 Q(x)=-1
x由一阶线性方程的求解公式
z?e?xdx2(??e2??xdx2dx?c)
=x?xc
y2?x?x2c
14
dyey?3x ?2dxxy两边同乘以e 令e
ydy(ey)2?3xeye?dxx2y
?z
dzdy ?eydxdx- 15 -
常微分方程(第三版,王高雄)课后答案
