第五章 常微分方程 一.?求解下列微分方程：? 1.? 0?1?1 = ÷ ? ? ? è ? - + ÷ ÷ ? ? ? ? è ? +? dy?y?x?e?dx?e? y?x?y?x? 解.? y?x?y?x?e?y?x?e?dy?dx + ÷ ? ? ? è ? - =?1? 1?.? 令? yu?x?u?y?x = =?,? .(将?y看成自变量)? dy?du?y?u?dy?dx + =? ,?所以? u?u? e?u?e?dy?du?y?u + - = +? 1?)?1?(? u?u?u?u?u? e?e?u?u?e?e?ue?dy?du?y + + - = - + - =? 1?1? y?dy?du?e?u?e?u?u - = + +?1? ,? y?dy?e?u?e?u?d?u?u - = + +?)?(? ,? y?y?c?e?u? u? 1?ln?ln?ln = - = ÷ ? ? ? è ? +? c?e?u?y? u + =?1? ,? y?x?u? e?y?x?c?e?u?c?y + = + =? ,? c?ye?x?y?x = ÷ ÷ ? ? ? ? è ? +? .? 2. ? ? ? í ì - = - + - - =? 1?)?1(? 2?2?'? 2?2? 2?2?y? x?xy?y? x?xy?y?y? 解.?令? xu?y?u?x?y = =?,? .? dx?du?x?u?dx?dy + =? ,?所以? 1?2?1?2?2?2 - + - - = +? u?u?u?u?dx?du?x?u? 1?2?1?1?2?1?2? 2? 2?3?2?2 - + - - - - = - - + - - =? u?u? u?u?u?u?u?u?u?u?dx?du?x? x?dx?du?u?u?u?u?u - = + + + - +? 1?1?2?2?3?2? x?dx?du?u?u?u - = ÷ ? ? ? è ? + + + -? 1?2?1?1?2? cx?u?u? ln?1?1?ln? 2 = + +? ,? cx?u?u = + +?1?1?2? .?由? 1?)?1?(?,?1?)?1?( - = - =? u?y? 得 所以?c=?0.?0?1?1? 2 = + +?u?u? ,?得到? 0?1= +?u? ,? 0?1= +?x?y? ,?即? x?y - =? .?二.?求解下列微分方程：? 1.? 2?1?2?2?2? sin?2?2?sin?'?1? x?e?y?x?y?y?x + + = + 解.?令? y?y?u?y?u? 2?sin?'?'?,?sin?2 = = 则? .?得到? 2?1?2?2? 2?'?1? x?e?xu?u?x + + = +? ,? 2?1?2?2? 1?1?2?'? 2?x?e?u?x?x?u? x + = + - + 为一阶线性方程 解得? |)?1?|?ln?(? 2?1?2? 2? x?x?c?e?u? x + + + = +? .?即? |)?1?|?ln?(?sin? 2?1?2?2? 2? x?x?c?e?y? x + + + = +? .?2.? 0?)?2?(? 2?2 = + - -? dx?y?dy?y?xy?x? 解.?原方程可化为? 2?2?1? y?x?y?x?dy?dx - + =? .? 即? 1?2?1?2 = ÷ ? ? ? è ? - +? x?y?y?dy?dx? ,?为一阶线性方程(y?为自变量,?x为因变量).? 解得:? y?e?cy?y?x? 1?2?2 + =? .?3.? 0?)?cos?1?(?cos?sin?ln?' = - +? y?x?y?y?x?xy? 解.?令? u?y =?cos?,?则? y?y?u? sin?'?' - =? .?原方程化为?0?)?1?(?ln?' = - + -? xu?u?x?x?u? x?u?x?x?u?u? ln?ln?'? 2 = + -? ,?为贝奴利方程.?x?u?x?x?u?u? ln?1?1?ln?1?'? 2 = × + -? .?令? u?z?1 =?,?则? 2?'?'? u?u?z - =? .?方程化为? x?z?x?x?z? ln?1?ln?1?' = +? ,?为一阶线性方程.?解得? x?c?x?z? ln?)?( + =? .?即? x?c?x?y? ln?cos?1 + =? ,? x?y?c?x? ln?cos?)?( = +? .? 三.?求解下列微分方程：?1.? 0?)?2?( = - +? dy?y?xe?dx?e? y?y? 解.? 0?2 = - +? ydy?dy?xe?dx?e? y?y? .?于是? 0?)?(? 2 = -dy?xe?d?y? .?所以方程解为? c?y?xe?y = -?2? .? 2.? 0?1?1? 2?2?2?2 = ÷ ÷ ? ? ? ? è ? - - + ÷ ÷ ? ? ? ? è ? - +? dy?x?y?y?x?dx?x?y?x? 解.? 0?1? 2?2?2?2 = - - - + +? dy?x?y?y?x?dx?x?y?dy?xdx? 设函数? )?,?(?y?x?u?满足? )?,?(?y?x?du? =? dy?x?y?y?x?dx?x?y? 2?2?2?2?1 - - -? .? 所以? 2?2?1?x?y?x?u - = ? ?? ,? )?(?arcsin?)?(?1?)?,?(? 2?2? y?y?x?y?dx?x?y?y?x?u j j + = + - = ò 所以? 2?2?2?2?2? )?(?'?1? x?y?y?x?y?y?x?y?x?y?u - - = + - - = ? ? j? .?于是? c?y?y = =? )?(?,?0?)?(?' j j 所以原方程的解为? c?y?x?y?x = + +? arcsin?2?1?2? 3.? 0?2?)?2?(? 2?2 = + + +? ydy?dx?x?y?x?解.?由原方程可得? 0?)?(?)?(? 2?2?2?2 = + + +? y?x?d?dx?y?x? 得到? 0?)?(? 2?2? 2?2 = + + +? y?x?y?x?d?dx? .?于是原方程解为? c?y?x?x = + +? )?ln(? 2?2? .? 四.?求解下列微分方程:?1.? )?1?(?2?'? 2 - - =? x?y?x?y?y? 解.? x?y?x?yy - = -? 2?)?1?(?'?2?令? u?y =?2? ,?得到? x?u?x?u - = -?)?1?(?'? 1?1?1?' - - = - -? x?x?u?x?u? 为一阶线性方程.?解得 ÷ ? ? ? è ? + - - - - =? c?x?x?x?x?u? )?1?ln(?1?)?1?(? .? 即? )?1?ln(?)?1?(?)?1?(?2 - - - + - =? x?x?x?x?c?y?2.? 6?3?'? y?x?y?xy = + 解.?该方程为贝奴利方程.?3?5?6?'? y?y?y?xy = + - -? .? 令? ,?5? u?y = -? '?'?5?6? u?y?y = - -? ,? 3?'?5? x?u?u?x = + -?2?5?5?'? x?u?x?u - = -? .?解得? )?2?5?(? 2?5 - + =? x?c?x?u? 于是? 3?5?5? 2?5?x?cx?y + = - 五.?设? )?(x y 在实轴上连续,? )?0?(?' y存在,?且具有性质? )?(?)?(?)?(? y?x?y?x y y y = +? ,?试求出? )?(x y? .?解.? )?0?(?)?0?(?)?0?0?( y y y = +? ,? )?0?(?)?0?(? 2 y y =? ,? 0?)?0?( = y? ,? 1?)?0?( = y? .?i)? 0?)?0?( = y? .?对于任何?x有? )?(?)?(?)?(? x?x?x?x D = D + y y y 所以? 0?)?0?(?)?(?)?(?lim?)?(?)?(?lim?)?(? 0?0 = = D = D + = Y ? D ? D y y y y y? x?x?x?x?x?x? x?x? . 所以? 0?)?( o?x y? .?ii)? 1?)?0?( = y? x?x?x?x?x?x?x?x?x?x?x?x?x?x D - = D - D = D - D = D - D +? )?0?(?)?(?)(?(?)?1)?(?)(?(?)?(?)?(?)?(?)?(?)?( y y y y y y y y y y 上式令? 0 ? Dx? ,?得到 ? í ì = =?1?)?0?(? )?0?(?'?)?(?)?(?'y y y y? x?x?解得? x?e?x? )?0?(?'?)?( y y =? .? 六.?求解下列方程:?1. ? í ì = = - +?1?)?0?(? 0?)?(?y? dy?x?y?ydx? 解.?可得 ? ? ? í ì = - = -?0?)?1?(? 1?x?y?x?dy?dx? .?这是以y?为自变量的一阶线性方程.? 解得? )?ln?(? y?c?y?x - =? .?0?)?1?( =?x? ,? 0 =?c? .?所以得解? y?y?x? ln - =? .? 2. ? ? ? í ì = = + + +?0?)2?(? 0?)?sin(?)?1'?( p?y? y?x?y?x? 解.?令? u?y?x = +? .?可得 ? ? ? í ì = = +? 2?)2?(? 0?sin?' p p?u u?xu? u?du?x?dx?sin = -? ,? )?cot?ln(csc?ln? u?u?x?c - =? ,? u?u?x?c? cot?csc - =? .?2?)2?( p p =?u? ,? 1?2?cot?2?csc? 2 = - = p p p?c? ,? 2 p =?c? .?解为? )?cot(?)?csc(?2? y?x?y?x?x + - + = p? .? 七.?求解下列方程:?1.? 0?1?)?'?(?'?'?)?1?(? 2?2 = + + +? y?y?x? 解.?令? dx?dp?y?p?y = =? '?'?,?'? 则? .?所以? 0?1?)?1?(? 2?2 = + + +? p?dx?dp?x? ,? 2?2? 1?1? x?dx?p?dp + - = +?c?x?p + - =?arctan?arctan? 所以? 1?tan?1? c?c?px?x?p = = - +? ,? px?c?c?x?p? 1?1 - = +? ,? x?c?x?c?p - = +? 1?1?)?1?( 于是? )?1?(? 1?1?1? 1?1?2?1?1?1?1? x?c?c?c?c?x?c?x?c?dx?dy + + + - = + - =? dx?x?c?c?c?c?dy ÷ ? ? ? è ? + + + - =? )?1?(? 1?1? 1?1?2?1?1? 解为? 2?1?2?1?2?1?1? |?1?|?ln?1?1? c?x?c?c?c?x?c?y + + + + - =? .? 2. ? í ì = = = - +? 1?)?2(?'?,?2?)?2(? 0?'?)?'?(?'?'? 2?y?y? y?y?x?xy?解.?令? dx?dp?y?p?y = =? '?'?,?'? 则?0?2 = - +? p?xp?dx?dp?x? ,? 2?p?x?p?dx?dp - = -? ,? 1?1?1?1? 2 - = -? p?x?dx?dp?p? 令? 1?)?2?(?1?'?1? 2 = - = =? u?dx?dp?p?u?u?p? ， ，则 于是得到? 1?1?' - = - -? u?x?u? ,? 1?1?' = +?u?x?u? 为?u?对于x?的一阶线性方程 解得? x?c?x?u + =?2?1? ,? 1?)?2?( =?u? ,?得?c=?0.? x?u?2?1 =? x?p?2?1?1 =? ,? x?dy?dx?2?1 =? ,? c?x?y + =?ln?2? ,? 2?ln?2?2?,?2?)?2?( - = =? c?y? 解得 所以? 2?)2?ln(?2?ln?2?2?ln?2? 2 + = - + =? x?x?y?3. ? í ì = = = +? 1?)?0(?'?,?2?)?0(? )?'?(?'?'?2? 2?y?y? y?y?y?解.?令? dy?dp?p?y?p?y = =? '?'?'?，则 得到? y?p?dy?dp?p = +?2?2? 令? u?p =?2? ,?得到? y?u?dy?du = + 为关于?y的一阶线性方程.?且? 1?)]?0?(?'?[?)?0?(?0?|? 2?2 = = = =? y?p?x?u? 解得? y?ce?y?u - + - =? 1?所以? 2?)?0?(? 1?2?1?)?0?(?0?|?1 - - + - = + - = = =? ce?ce?y?x?u? y? ,? 0 =?c? .? 于是? 1 - =?y?u? ,? 1 - ± =? y?p?dx? y?dy ± = -1? ,? 1?1?2? c?x?y + ± = -? ,? 2?2?1? 1?c?x?y + ± = -?2?)?0?( =?y? ,?得到? 1?2? 1 =?c? ,?得解? 1?2?1 + ± = -? x?y 八.?求解下列微分方程:?1.? 0?'?'?'?2?'?'?'?2?)?4?(?)?5?( = + + + + +? y?y?y?y?y?y? 解.?特征方程? 0?1?2?2? 2?3?4?5 = + + + + + l l l l l?0?)?1?)(?1?(? 2?2 = + + l l? i?i - = = - =? 5?,?4?3?,?2?1? ,?,?1 l l l 于是得解? x?x?c?c?x?x?c?c?e?c?y? x? cos?)?(?sin?)?(? 5?4?3?2?1 + + + + = -? 2. ? í ì - = = = = = - + -? 14?)?0(?'?'?'?,?6?)?0(?'?'?,?0?)?0(?'?,?1?)?0(? 0?6?'?10?'?'?5?)?4?(? y?y?y?y? y?y?y?y? 解.?特征方程? 0?6?10?5?2?4 = - + - l l l? ,? 0?)?2?2?)(?3?)(?1?(? 2 = + - + - l l l l?1? 1 = l? ,? 3?2 - = l? ,? i ± =1?4?,?3 l 得通解为? )?sin?cos?(? 4?3?3?2?1? x?c?x?c?e?e?c?e?c?y? x?x?x + + + = - 由? 14?)?0?(?'?'?'?,?6?)?0?(?'?'?,?0?)?0?(?'?,?1?)?0?( - = = = =? y?y?y?y? 得到? 2?1?1 - =?c? ,? 2?1?2 =?c? ,? 1?3 =?c? ,? 1?4 =?c?得特解? )?sin?(cos?2?1?2?1? 3? x?x?e?e?e?y? x?x?x + + + - = - 九.?求解下列微分方程:?1.? x?x?x?y?y? cos?2?2?sin?3?'?' + + = + 解.?特征方程? 0?1?2 = + l? ,? i ± = l 齐次方程通解? x?c?x?c?y? sin?cos? 2?1 + = 非齐次方程特解:? x?x?D?y = + =? 1?1? 2?*?1? x?x?x?D?x?D?y? 2?sin?2?sin?1?4?3?2?sin?1?1?3?2?sin?3?1?1? 2?2?*?2 - = + - = + = + =? x?D?y? cos?2?1?1? 2?*?3 + = 考察? 1?2?1?1?2?1?2?1?2?11?)?(? 1?2?2?1?1? 2?2?2? D?i?D?e?iD?D?e?i?D?e?e?D? ix?ix?ix?ix + = + = + + = +? =? )?)(?sin?(cos?1? 2?1?1?2?1?)4?2?1?(?1?2? ix?x?i?x?xe?i?iD?e?D?iD?e? ix?ix?ix - + = = = +?=? x?ix?x?x? cos?sin - 所以? x?x?x? D?y? sin?cos?2?1?1?2?*?3 = + = 所以通解为? x?x?x?x?x?c?x?c?y? sin?2?sin?sin?cos? 2?1 + - + + =? 2. ? í ì = = + = +? 0?)?0(?'?)?0(? sin?4?2?'?'? y?y? x?xe?y?y? x 解.?特征方程? 0?1?2 = + l? ,? i ± = l 齐次方程特解? x?c?x?c?y? sin?cos? 2?1 + = 非齐次方程通解? x?D?D?e?x?D?e?xe?D?y? x?x?x? 2?2?1?2?1?)?1?(? 1?2?2?1?1? 2?2?2?*?1 + + = + + = + =? =? )?1?(?2?1?2?1?2 - = ÷ ? ? ? è ? -? x?e?x?D?e? x?x? x?x?x? D?y? cos?2?sin?4?1?1?2?*?2 - = + =? (计算方法同上题,?取? ix?e?D? 1?1?2 + 的虚部)?所以? x?x?x?e?x?c?x?c?y? x? cos?2?)?1?(?sin?cos? 2?1 - - + + = 由? 0?)?0?(?'?)?0?( = =?y?y? 可得? 2?,?1?2?1 = =?c?c?得解? x?x?x?e?x?x?y? x? cos?2?)?1?(?sin?2?cos - - + + =?3.? ax?e?y?y?y = + +? 4?'?4?'?'?解.?特征方程? 0?4?4? 2 = + + l l? ,? 2?2?,?1 - = l? x?e?x?c?c?y? 2?2?1? )?( - + =?i)?2 - =?a? x?x?x? e?x?D?e?e?D?y? 2?2?2?2?2?2?*? 2?1?1?)?2?2?(? 1?)?2?(?1 - - - = + - = + =? ii)? 2 - 1?a? 2?2?*? )?2?(?)?2?(?1 + = + =? a?e?e?D?y? ax?ax? 所以 ? ? ? ? í ì + + + + + = - - -? x?x? ax?x? e?x?e?x?c?c? e?a?e?x?c?c?y? 2?2?2?2?1? 2?2?2?1? 2?1?)?(? )?2(?1?)?(? 2?2- = - 1?a?a? 十.?求解下列微分方程:?1.? )?sin(ln?2?'?'?'?2? x?y?xy?y?x = + + 解.?令? x?t?e?x?t? ln?, = = 则 得? dt?dy?dt?y?d?y?x?dt?dy?xy - = =? 2?2?2? '?'?'? 得到方程? t?y?y? sin?2?'?' = +? .?解得? t?t?t?c?t?c?y? cos?sin?cos? 2?1 - + = 所以得解? x?x?x?c?x?c?y? ln?cos?ln?ln?sin?ln?cos? 2?1 - + = 2.? )?1?ln(?)?1?(?6?'?)?1?(?'?'?)?1?(? 2 + + = + + - +? x?x?y?y?x?y?x?解.?令? )?1?ln(?,?1 + = = +? x?t?e?x? t?则 得? dt?dy?dt?y?d?y?x? dt?dy?y?x - = + = +? 2?2?2? '?'?)?1?(? '?)?1?(? 得到方程? t?te?y?y?y? 6?'?2?'?' = + -? .?解得? t?t? e?t?e?t?c?c?y? 3?2?1? )?( + + = 所以得解? )?1?(?ln?)?1?(?)?1?)(?1?ln(?(? 3? 2?1 + + + + + + =? x?x?x?x?c?c?y?十一.一质量为?m?的物体,?在粘性液体中由静止自由下落,?假如液体阻力与运动速度成正比,? 试求物体运动的规律.?解.?取物体的初始位置为坐标原点,x坐标向下为正向.?并以? )?(t?x?表示在时刻t时的物体位置.? 物体所受的重力为mg,?阻力为?dt?dx?k? (k?为比例系数).?由牛顿定律得到: ? ? ? í ì = = = -? 0?)?0?(?'?)?0?(? 2?2?x?x? dt?x?d?m?dt?dx?k?mg? .?即 ? ? ? í ì = = = +? 0?)?0?(?'?)?0?(? '?'?'? x?x? g?x?m?k?x? 解得? t?k?mg?e?c?c?x? t?m?k + + = -?2?1? 于是? k?mg?e?c?m?k?x? t?m?k + - = -?2?'0?)?0?( =?x? ,?得到? 2?1? c?c - =?k?mg?c?m?k?x + - = =? 2?)?0(?'?0?所以? 2?2?2? k?g?m?c =? ,? 2?2?2?1? k?g?m?c?c - = - = 所求解为? t?k?mg?e?k?g?m?k?g?m?x? t?m?k + + - = -?2?2?2?2? .? C?A O