高数同济五版(7)

习题12?4

1? 求下列微分方程的通解? (1)

dy?y?e?x? dx?dxdx 解 y?e?(e?x?e?dx?C)?e?x(e?x?exdx?C)?e?x(x?C)?

?? (2)xy??y?x2?3x?2?

解 原方程变为y??1y?x?3?2xx?

1 y?e??1xdx[?(x?3?2?xdxx)?edx?C] ?1x[?(x?3?21x)xdx?C]?x[?(x2?3x?2)dx?C] ?11332x(3x?2x?2x?C)?13x2?3C2x?2?x? (3)y??ycos x?e?sin x?

解 y?e??cosdx(?e?sinx?e?cosxdxdx?C)

?e?sixn(?e?sixn?esinxdx?C)?e?sixn(x?C)?

(4)y??ytan x?sin 2x?

解 y?e??tanxdx(?sin2x?e?tanxdxdx?C)

?elncosx(?sin2x?e?lncoxsdx?C)

?cosx(?2sinxcosx?1cosxdx?C) ?cos x(?2cos x+C)?C cos x?2cos2x ? (5)(x2?1)y??2xy?cos x?0? 解 原方程变形为y??2xx2?1y?cosxx2?1? ??2xdx2x y?ex2?1(?cosx?2?1dxx2?1?exdx?C) ?1cosx21x2?1[?x2?1?(x?1)dx?C]?x2?1(sinx?C)? (6)

d??3??2? d??3d?3d? 解 ??e?(?2?e?d??C)

?e?3?(?2e3?d??C)

?e?3?(2e3??C)?2?Ce?3??

33 (7)

dy?2xy?4x? dx?2xdx2xdx 解 y?e?(?4x?e?dx?C)

?e?x(?4x?exdx?C) ?e?x(2ex?C)?2?Ce?x? (8)yln ydx?(x?ln y)dy?0? 解 原方程变形为

??1dyylny22222dx11?x?? dyylnyy1 x?e ? ?dy1?(??eylnydy?C) y11(??lnydy?C) lnyy1121C(lny?C)?lny?? lny22lny (9)(x?2)dy?y?2(x?2)3? dxdy1?y?2(x?2)2? dxx?22??1dxx?2dx 解 原方程变形为 y?e?x?2dx1[?2(x?2)?e?C]

?(x?2)[?2(x?2)2?2

1dx?C] x?23

?(x?2)[(x?2)?C]?(x?2)?C(x?2)?

dy?2y?0? (10)(y2?6x)dx 解 原方程变形为

?ydy3dx31?x??y? dyy23 x?e??dy1[?(?y)?eydy?C]

2 ?y3(? ?y3(12?y?1dy?C) y311?C)?y2?Cy3? 2y2 2? 求下列微分方程满足所给初始条件的特解? (1)

dy?ytanx?secx? y|x?0?0? dxtanxdx?tanxdx(?secx?e?dx?C)

解 y?e? ?11(?secx?cosxdx?C)?(x?C)? cosxcosx 由y|x?0?0? 得C?0? 故所求特解为y?xsec x ? (2)

dyysinx? y|x???1? ??dxxx1??dxex( 解 y? ?sinx?xdx?x?edx?C)

11sinx1(??xdx?C)?(?cosx?C)? xxx1(??1?cosx)? x 由y|x???1? 得C???1? 故所求特解为y? (3)

dy?ycotx?5ecosx? y|???4?

x?dx2?cotxdxcotxdx(?5ecosx?e?dx?C) 解 y?e? ? 由y| (4)

11xx(?5ecos?sinxdx?C)?(?5ecos?C)? sinxsinxx????4? 得C?1? 故所求特解为y?21(?5ecosx?1)? sinxdy?3y?8? y|x?0?2? dx?3dx3dx 解 y?e?(?8?e?dx?C)

?e?3x(8?e3xdx?C)?e?3x(8e3x?C)?8?Ce?3x?

33 由y|x?0?2? 得C??2? 故所求特解为y?2(4?e?3x)?

33dy2?3x2 (5)?y?1? y|x?1?0?

dxx3 解 y?e ???2?3x2dxx3(?2?3x2?3dx1?exdx?C)

1113x2xe?21?x23x21(?3edx?C)?xe(ex?C)?

2x1?111 由y|x?1?0? 得C??? 故所求特解为y?x3(1?ex2)?

2e21 3? 求一曲线的方程? 这曲线通过原点? 并且它在点(x? y)处的切线斜率等于2x?y ? 解 由题意知y??2x?y? 并且y|x?0?0? 由通解公式得

dx?dx y?e?(?2xe?dx?C)?ex(2?xe?xdx?C)

?ex(?2xe?x?2e?x?C)?Cex?2x?2?

由y|x?0?0? 得C?2? 故所求曲线的方程为y?2(e?x?1)?

4? 设有一质量为m的质点作直线运动? 从速度等于零的时刻起? 有一个与运动方向一至、大小与时间成正比(比例系数为k1)的力作用于它? 此外还受一与速度成正比(比例系数为k2)的阻力作用? 求质点运动的速度与时间的函数关系? 解 由牛顿定律F?ma? 得m 由通解公式得 v?e??k2mdtx

kdvk2dv?v?1t? ?k1t?k2v? 即

dtmmdt(?k1mt?e?mdtk2dt?C)?ek2?k2mt(?k1mk2t?emdt?C)

t ?e?k2mt(k1k2k2temt?k1m2k2emt?C)?

由题意? 当t?0时v?0? 于是得C?k1m2k2? 因此

v?e?k2mt(k1k2k2temt?k1m2k2k2mk2em?tk1m2k2)

即 v?k1k2t?k1m2k2(1?e?t)?

5? 设有一个由电阻R?10?、电感L?2h(亨)和电源电压E?20sin5t V(伏)串联组成的电路? 开关K合上后? 电路中有电源通过? 求电流i与时间t的函数关系? 解 由回路电压定律知 20sin5t?2 由通解公式得

?5dt5dt i?e?(?10sin5t?e?dt?C)?sin5t?cos5t?Ce?5t?

didi?10i?0? 即?5i?10sin5t? dtdt 因为当t?0时i?0? 所以C?1? 因此

i?sin5t?cos5t?e?5t?e?5t?2sin5(t?

6? 设曲?yf(x)dx?[2xf(x)?x2]dy在右半平面(x?0)内与路径无关? 其中f(x)可导? 且

L?4)(A)?

f(1)?1? 求f(x)?

解 因为当x?0时? 所给积分与路径无关? 所以

??[yf(x)]?[2xf(x)?x2]? ?y?x即 f(x)?2f(x)?2xf?(x)?2x? 或 f?(x)?1f(x)?1? 2x??1dx2x(因此 f(x)?e?1?e?2xdx1dx?C)?1x(?xdx?C)?2Cx?? 3x由f(1)?1可得C?211? 故f(x)?x??

333x 7? 求下列伯努利方程的通解? (1)

dy?y?y2(cosx?sinx)? dx

解 原方程可变形为

d(y?1)1dy1 2?y?1?sinx?cosx? ??cosx?sinx? 即

dxydxydx?dx y?1?e?[?(sinx?cosx)?e?dx?C]

?e?x[?(cosx?sinx)exdx?C]?Cex?sinx? 原方程的通解为 (2)

1?Cex?sinx? ydy?3xy?xy2? dx 解 原方程可变形为

d(y?1)1dy1? 即?3xy?1??x? ?3x?x2dxyydx?3xdx3xdx y?1?e?[?(?x)?e?dx?C]

?e ?e?32x2(??xe32x2dx?C)

32?32x2(??x12x1e?C)?Ce2?? 3332?x11原方程的通解为?Ce2??

y332 (3)

dy11?y?(1?2x)y4? dx33 解 原方程可变形为

d(y?3)1dy111?y?3?2x?1? 4??(1?2x)? 即3dx3ydx3ydx?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

1?Cex?2x?1? 3y (4)

dy?y?xy5? dx 解 原方程可变形为

d(y?4)1dy1 5?4y?4??4x? ?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

11?4x? ??x??Ce4y4

(5)xdy?[y?xy3(1?ln x)]dx?0? 解 原方程可变形为

d(y?2)2?21dy11 3???2?(1?lnx)? 即?y??2(1?lnx)?

dxxydxxy y?2?2??dxex[?2?(1?lnx)?e?xdx2dx?C]

? ?1[?2?(1?lnx)x2dx?C] 2xC24?xlnx?x?

9x231C24原方程的通解为2?2?xlnx?x?

39yx 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

x

dv?vvf(v)dx? ??22xxg(v)即 积分得

g(v)1du?dx?

v[g(v)?f(v)]xg(v)?v[g(v)?f(v)]du?lnx?C?

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?(x?y)2? dx 解 令u?x?y? 则原方程化为

(1)

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)? (2)

dy1??1? dxx?y 解 令u?x?y? 则原方程化为 1?du1??1? 即dx??udu? dxu12两边积分得

x??u2?C1?

将u?x?y代入上式得原方程的通解

x??(x?y)2?C1? 即(x?y)2??2x?C(C?2C1)? (3)xy??y?y(ln x?ln y)?

解 令u?xy? 则原方程化为 x(1duuuu11?2)??lnu? 即dx?du? xdxxxxxulnuCx

12两边积分得

ln x?ln C?lnln u? 即u?e? 将u?xy代入上式得原方程的通解 xy?eCx? 即y?1Cxe? 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)?cos x? 令u?y?sin x?1? 则原方程化为

du?cosx?u2?cosx? 即12du?dx?

dxu2

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 ?

(5)y(xy?1)dx?x(1?xy?xy)dy?0 ? 解 原方程变形为

dyy(xy?1)? ??22dxx(1?xy?xy)22

1?x?C? 即y?1?sinx?1?

y?sinx?1x?C令u?xy? 则原方程化为

u(u?1)1duu1duu3? 即? ?2??2?222xdxx(1?u?u)xdxxx(1?u?u)分离变量得

1111dx?(3?2?)du? xuuu11??lnu? 2u2u12xy22两边积分得 lnx?C1??将u?xy代入上式得原方程的通解 lnx?C1???1?lnxy? xy即 2x2y2ln y?2xy?1?Cx2y2(C?2C1)?

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