7? 求下列伯努利方程的通解?
dy (1)?y?y2(cosx?sinx)?
dx 解 原方程可变形为
d(y?1)dy11?y?1?sinx?cosx? 2??cosx?sinx? 即dxydxydx?dx y?1?e?[?(sinx?coxs)?e?dx?C]
?e?x[?(cosx?sinx)exdx?C]?Cex?sinx? 原方程的通解为1?Cex?sinx?
ydy?3xy?xy2? dx 解 原方程可变形为 (2)
dyd(y?1)11?3xy?1??x? 2?3x?x? 即
dxyydx?3xdx3xdx y?1?e?[?(?x)?e?dx?C]
?3x2?e2(??3x2xe2dx?C)
?3x2?C)?Ce23x2?3x21?e2(?e23?1?
3?x211原方程的通解为?Ce2?? y33dy11?y?(1?2x)y4? dx33 解 原方程可变形为 (3)
d(y?3)?31dy?11?1(1?2x)?y?2x?1? 4? 即33dxdx3yydx?dx y?3?e?[?(2x?1)?e?dx?C]
?ex[?(2x?1)e?xdx?C]??2x?1?Cex? 原方程的通解为13?Cex?2x?1?
ydy?y?xy5? dx 解 原方程可变形为 (4)
dy1d(y?4)1?4y?4??4x? 5?4?x? 即
dxydxy?4dx4dx y?4?e?[?(?4x)?e?dx?C]
?e?4(?4?xe4xdx?C) ??x?1?Ce?4x?
4原方程的通解为14??x?1?Ce?4x?
4y
(5)xdy?[y?xy3(1?ln x)]dx?0? 解 原方程可变形为
dy11d(y?2)2?21?y??2(1?lnx)? 3???2?(1?lnx)? 即
dxxydxxy y?2??2dx?ex[?2?xdxdx?C] (1?lnx)?e?2 ?1[?2?(1?lnx)x2dx?C] 2x2xlnx?4x? ?C?9x232xlnx?4x? ?原方程的通解为12?C9yx23 8? 验证形如yf(xy)dx?xg(xy)dy?0的微分方程? 可经变量代换v?xy化为可分
离变量的方程? 并求其通解? 解 原方程可变形为
dy?yf(xy)? ? dxxg(xy) 在代换v?xy下原方程化为
xdv?vvf(v) dx? ??22xxg(v)g(v)du?1dx? 即
v[g(v)?f(v)]xg(v)du?lnx?C? 积分得 ?v[g(v)?f(v)]对上式求出积分后? 将v?xy代回? 即得通解?
9? 用适当的变量代换将下列方程化为可分离变量的方程? 然 后求出通解?
dy (1)?(x?y)2?
dx 解 令u?x?y? 则原方程化为 du?1?u2? 即dx?du2?
dx1?u两边积分得
x?arctan u?C?
将u?x?y代入上式得原方程的通解
x?arctan(x?y)?C? 即y??x?tan(x?C)?
dy (2)?1?1?
dxx?y 解 令u?x?y? 则原方程化为 1?du?1?1? 即dx??udu?
dxu两边积分得
x??1u2?C1?
2将u?x?y代入上式得原方程的通解
x??1(x?y)2?C1? 即(x?y)2??2x?C(C?2C1)?
2 (3)xy??y?y(ln x?ln y)?
解 令u?xy? 则原方程化为
x(1du?u2)?u?ulnu? 即1dx?1du?
xdxxxxxulnu两边积分得
ln x?ln C?lnln u? 即u?eCx? 将u?xy代入上式得原方程的通解
xy?eCx? 即y?1eCx?
x (4)y??y2?2(sin x?1)y?sin2x?2sin x?cos x?1? 解 原方程变形为
y??(y?sin x?1)2?cos x? 令u?y?sin x?1? 则原方程化为 du?cosx?u2?cosx? 即1du?dx? 2dxu两边积分得
?1?x?C?
u将u?y?sin x?1代入上式得原方程的通解
1?x?C? 即y?1?sinx?1? ?y?sinx?1x?C
(5)y(xy?1)dx?x(1?xy?x2y2)dy?0 ? 解 原方程变形为
dyy(xy?1)??? 22dxx(1?xy?xy)令u?xy? 则原方程化为
3u(u?1)1duu1duu???2? ? 即? xdxx2xdxx2(1?u?u2)x(1?u?u2)分离变量得
1dx?(1?1?1)du? 32xuuu两边积分得
lnx?C1??12?1?lnu?
2uu将u?xy代入上式得原方程的通解
1?lnxy? ? lnx?C1??12x2y2xy即 2x2y2ln y?2xy?1?Cx2y2(C?2C1)?