习题12?4
1? 求下列微分方程的通解?
dy (1)?y?e?x?
dx 解 y?e??dx(?e?x?e?dxdx?C)?e?x(?e?x?exdx?C)?e?x(x?C)? (2)xy??y?x2?3x?2?
解 原方程变为y??1xy?x?3?2x?
1
y?e??1xdx[?(x?3?2)?e?dxxxdx?C]
?1x[?(x?3?2x)xdx?C]?1x[?(x2?3x?2)dx?C]
?1x(13x3?32x2?2x?C)?13x2?32x?2?Cx?
(3)y??ycos x?e?sin x?
解 y?e??cosdx(?e?sinx?e?cosxdxdx?C)
?e?sixn(?e?sixn?esixndx?C)?e?sixn(x?C)? (4)y??ytan x?sin 2x?
解 y?e??tanxdx(?sin2x?e?tanxdxdx?C)
?elncosx(?sin2x?e?lncoxsdx?C) ?coxs(?2sinxcoxs?1coxsdx?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? x
y?e??2x2?1dx(?coxs?e?2xx2?1x2?1dxdx?C)
?1oxsx2?1[?cx2?1?(x2?1)dx?C]?1x2?1(sixn?C)?
(6)
d??3??2? d? 解 ??e??3d?(?2?e?3d?d??C) ?e?3?(?2e3?d??C) ?e?3?(2e3??C)?2?Ce?3??
33dy (7)?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dx?1x?1?
dyylnyy
??1dyx?eylny(222221?e?ylnydydy?C) ?y1 ?1(?1?lnydy?C)
lnyy ?1(1ln2y?C)?1lny?C?
lny22lnydy?y?2(x?2)3? dxdy 解 原方程变形为?1y?2(x?2)2?
dxx?2 (9)(x?2)dx?dx y?e?x?2[?2(x?2)2?e?x?2dx?C]
11 ?(x?2)[?2(x?2)2?1dx?C]
x?2 ?(x?2)[(x?2)2?C]?(x?2)3?C(x?2)?
dy (10)(y2?6x)?2y?0?
dx 解 原方程变形为dx?3x??1y?
dyy2??dy?ydy1 x?e[?(?y)?eydy?C] 233 ?y3(?1?y?13dy?C)
2y ?y3(1?C)?1y2?Cy3? 2y2 2? 求下列微分方程满足所给初始条件的特解?
dy (1)?ytanx?secx? y|x?0?0?
dx 解 y?e?tanxdx?tanxdx(?secx?e?dx?C)
?1(?secx?coxsd?xC)?1(x?C)? coxscoxs 由y|x?0?0? 得C?0? 故所求特解为y?xsec x ?
dyy (2)??sinx? y|x???1?
dxxx 解
??1dxy?ex(sinx?e?xdxdx?C) ?x1x?xd? ?1(?sinxC)?1(?coxs?C)? xxx 由y|x???1? 得C???1? 故所求特解为y?1(??1?cosx)?
x (3)
dy?ycotx?5ecosx? y|x????4? dx2?cotxdxcotxdx 解 y?e?(?5ecosx?e?dx?C)
?1(?5ecoxs?sinxd?xC)?1(?5ecoxs?C)?
sinxsinx 由y|???4? 得C?1? 故所求特解为y?1(?5ecosx?1)?
x?sinx2 (4)
dy?3y?8? y|x?0?2? dx 解 y?e??3dx(?8?e?3dxdx?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?3x2y?1? y|x?1?0? (5)?dxx3 解
xdx??2?33y?ex(2?1?e?12?3x2dxx3dx?C)
13x2?xe(1e?x2dx?C)?x3ex2(1e?x2?C)? ?x32111?1 由y|x?1?0? 得C??1? 故所求特解为y?1x3(1?ex2)?
22e 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(ex?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?
kdvk 解 由牛顿定律F?ma? 得mdv?k1t?k2v? 即?2v?1t?
dtmmdt 由通解公式得 v?e??k2dtm(22dt?2tk1tk1?mmmt?edt?C)?e(t?e?m?mdt?C)
kkk ?e?k2mt22tk1mtk1m(te?2em?C)? k2k2kk 由题意? 当t?0时v?0? 于是得C?k2k2ktk1tk1m2tk1mm(tem?em?2)2k2k2k2kk1m? 因此 2k2 v?e?
?2tk1k1m即 v?t?2(1?em)?
k2k2 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?2di?10i?0? 即di?5i?10sin5t?
dtdt 由通解公式得
?5dt5dt i?e?(?10sin5t?e?dt?C)?sin5t?cos5t?Ce?5t?
因为当t?0时i?0? 所以C?1? 因此
i?sin5t?co5st?e?5t?e?5t?2sin5t(??)(A)?
4
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)
L可导? 且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因此
??1dxf(x)?e2x(?2xdxdx?C)?1(xdx?C)?2x?C? 1?e?3x?x121由f(1)?1可得C?1? 故f(x)?x??
33x3