.*
习题8.1
1.指出下列微分方程的阶数，并指出哪些方程是线性微分方程： (1)x(y?)?2yy??xy?0 (2) xy?xy??y?0 (3)xy????4y???(sinx)y?0 (4)解 (1) 1阶 非线性 (2) 1阶 线性 (3) 3阶 线性 (4) 1阶 线性
2.验证下列函数是否是所给微分方程的解 (1) xy??y?cosx,y?222dp?p?sin2? d?sinx x (2) (1?x2)y??xy?2x,y?2?C1?x2 (C为任意常数) (3) y???2y??y?0,y?Ce (C为任意常数) (4) y???(?1??2)y???1?2y?0,y?C1e22x?1x?C2e?2x (C1 ,C2为任意常数)
(5) (x?2y)y??2x?y,x?xy?y?C (C为任意常数) (6) (xy?x)y???xy??yy??2y??0,y?ln(xy) 解 (1) 是，左=x2xcosx?sinxsinx??cosx=右
xx22 (2) 是，左=(1?x)?Cx1?x2x?x(2?C1?x2)?2x=右
x (3) 是，左=Ce?2Ce?Ce?0=右 (4) 是，左=
?2x2?1x(C1?1e?C2?2)?(?1??2)(C1?1e?1x?C2?2e?2x)??1?2(C1e?1x?C2e?2x)?0 2ex=右
(5) 是，左=(x?2y)2x?y?2x?y?右
x?2y2xy2?xy3?2xyy2yy?x?y?2 (6) 是，左=(xy?x) 32xy?xxy?x(xy?x)(xy?x).*
2xy2?xy3?2xyxy2(y2?2y)(xy?x)???0 =
(xy?x)2(xy?x)2(xy?x)2 = 右
3.求下列微分方程的解
d2ydy?cosx; (1) ?2; (2) 2dxdx(1?y2)x (3) (1?y)dx?(1?y)dy?0 (4) y?? 2(1?x)y解 (1) (2)
?dy??2dx,y?2x?C ?y??dx??cosxdx,y??sinx?C11
?y?dx??(sinx?C)dx,y??cosx?Cx?C12
(3)
1?y?1?ydy??dx
解得
?(1?y)?2?1?ydy??dx
??dy??2dy??dx 1?y即 ?y?2ln|1?y|?x?C
(4)
yxdy??1?y2?(1?x2)dx
222 解得 ln(1?y)?ln(1?x)?C1
1?y22?C整理得
1?x24.已知曲线y?f(x)经过原点，并且它在点(x,y)处的切线的斜率等于2x，试求这条曲线的方程。 解 已知 y??2x 解得 y?2223x?C 3 又知曲线过原点，得C?0 所求曲线方程为y?23x 3.*
习题8.2
1.用分离变量法求下列微分方程的解
(1) y??4xy (2) xy??ylny?0 (3) y??10x?y (4) secxtanydx?secytanxdy?0
22 (5)
xydx?dy?0,y|x?0?1 (6) y??e2x?y,y|x?0?0 1?y1?x解 (1)
??1ydy??4xdx 解得 y?(x2?C)2
(2)
dydx?? 解得 y?eCx ylnyx (3)
?yxx?y?yx?10?10?C10?10?C 解得 即 10dy?10dx??sec2ysec2xdy???dx 解得 ln|tany|??ln|tanx|?C1 (4) ?tanytanx 整理得 tanx?tany?C (5)
?y(1?y)dy??x(1?x)dx 解得
5 612131213y?y?x?x?C 2323 由于 y|x?0?1 ，解得 C? 则
121312135y?y?x?x? 2323612xe?C 2 (6)
?y?y2x?e? 解得 edy?edx?? 由于 y|x?0?0 则 C?? 原方程解为 2e?y3 2?3?e2x
2.求下列齐次方程的解 (1) xy??ylndyx?yy? (2) dxx?yx (3) xy??y?y2?x2?0 (4) x2dy?(y2?xy?x2)dx
.*
(5) y?x22dydy?xy (6) x(x?2y)y??y2?0,y|x?1?1 dxdxy，代入方程得 xu?xdu?ulnu dx解 (1) 令u?分离变量得
dudx?
u(lnu?1)x两边积分得
ln|lnu?1|?ln|x|?C1
整理得 |lnu?1|?C2|x| 将u?y回代，即得原方程通解 xlny?1?Cx xydyx (2) 原式可化为 ?ydx1?x1?令u?y，代入方程得 xu?xdu1?u? dx1?u分离变量得
(1-u)dudx? x1?u2两边积分得
1arctanu?ln(1?u2)?ln|x|?C12 将u?y回代，即得原方程通解 x.*
yy22arctan?ln(1?2)?lnx2?C
xx整理得 2arctany?ln(x2?y2)?C x2dyy?y??????1 (3) 原式可化为
dxx?x?令u?y，代入方程得 xxdu?u2?1 dx分离变量得
duu2?1两边积分得
?dx xln|u?u2?1|?ln|x|?C1
即 |u?u2?1|?C|x| 将u?y回代，即得原方程通解 xy?y?????1?Cx x?x?2dy?y?y(4) 原式可化为 ?????1
dx?x?x令u?2y，代入方程得 xu?xdu?u2?u?1 dx分离变量得
dudx?
u2?2u?1x两边积分得
微积分习题集讲解与规范标准答案
