最新常微分方程(第三版)答案

常微分方程(第三版)

答案

常微分方程习题答案

2.1

1.?Skip Record If...?,并求满足初始条件：x=0,y=1的特解.

解：对原式进行变量分离得

?Skip Record If...??Skip Record If...?并求满足初始条件：x=0,y=1的特解. 解：对原式进行变量分离得：

?Skip Record If...?3 ?Skip Record If...?

解：原式可化为：

?Skip Record If...??Skip Record If...??Skip Record If...?

?Skip Record If...?

12．?Skip Record If...?

解?Skip Record If...??Skip Record If...?

?Skip Record If...?

15．?Skip Record If...?

?Skip Record If...?16．?Skip Record If...?

解：?Skip Record If...?

?Skip Record If...?，这是齐次方程，令?Skip Record If...?

17. ?Skip Record If...?

解：原方程化为?Skip Record If...?

令?Skip Record If...?

方程组?Skip Record If...??Skip Record If...?

则有?Skip Record If...?

令?Skip Record If...?

当?Skip Record If...?当?Skip Record If...?

另外

?Skip Record If...?

?Skip Record If...?

19. 已知f(x)?Skip Record If...?.

解：设f(x)=y, 则原方程化为?Skip Record If...?两边求导得?Skip Record If...?

?Skip Record If...??Skip Record If...??Skip Record If...?

?Skip Record If...??Skip Record If...?

20.求具有性质 x(t+s)=?Skip Record If...?的函数x(t),已知x’(0)存在。

解：令t=s=0 x(0)=?Skip Record If...?=?Skip Record If...?若x(0)?Skip Record If...?0 得x?Skip Record If...?=-1矛盾。

所以x(0)=0. x’(t)=?Skip Record If...?)

?Skip Record If...??Skip Record If...?两边积分得arctg x(t)=x’(0)t+c 所以

x(t)=tg[x’(0)t+c] 当t=0时 x(0)=0 故c=0 所以

x(t)=tg[x’(0)t]

习题2.2

求下列方程的解

1．?Skip Record If...?=?Skip Record If...?

解： y=e ?Skip Record If...?(?Skip Record If...?e?Skip Record If...??Skip Record If...?) =e?Skip Record If...?[-?Skip Record If...?e?Skip Record If...?(?Skip Record If...?)+c]

=c e?Skip Record If...?-?Skip Record If...? (?Skip Record If...?)是原方程的解。

2．?Skip Record If...?+3x=e?Skip Record If...?

解：原方程可化为：?Skip Record If...?=-3x+e?Skip Record If...?

所以：x=e?Skip Record If...? (?Skip Record If...?e?Skip Record If...? e?Skip Record

If...??Skip Record If...??Skip Record If...?)

=e?Skip Record If...? (?Skip Record If...?e?Skip Record If...?+c)

=c e?Skip Record If...?+?Skip Record If...?e?Skip Record If...?是原方程的解。3．?Skip Record If...?=-s?Skip Record If...?+?Skip Record If...??Skip Record If...?

解：s=e?Skip Record If...?(?Skip Record If...?e?Skip Record If...??Skip Record If...? )

=e?Skip Record If...?(?Skip Record If...?)

= e?Skip Record If...?(?Skip Record If...?)

=?Skip Record If...?是原方程的解。

4．?Skip Record If...??Skip Record If...?， n为常数.

解：原方程可化为：?Skip Record If...??Skip Record If...?

?Skip Record If...?

?Skip Record If...?是原方程的解.

5．?Skip Record If...?+?Skip Record If...?=?Skip Record If...?

解：原方程可化为：?Skip Record If...?=-?Skip Record If...?

?Skip Record If...?(?Skip Record If...?)

?Skip Record If...??Skip Record If...?

=?Skip Record If...?是原方程的解.

6．?Skip Record If...??Skip Record If...?

解：?Skip Record If...??Skip Record If...?

=?Skip Record If...?+?Skip Record If...?

令?Skip Record If...??Skip Record If...?则?Skip Record If...??Skip Record

If...?=u?Skip Record If...?

因此：?Skip Record If...?=?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

?Skip Record If...?（*）

将?Skip Record If...??Skip Record If...?带入（*）中得：?Skip Record If...?是原方程的解.

?Skip Record If...?

?Skip Record If...??Skip Record If...?

?Skip Record If...??Skip Record If...??Skip Record If...??Skip Record If...??Skip Record If...??Skip Record If...??Skip Record If...?

?Skip Record If...??Skip Record If...?

13

?Skip Record If...?

这是n=-1时的伯努利方程。

两边同除以?Skip Record If...?，

?Skip Record If...?

令?Skip Record If...??Skip Record If...?

?Skip Record If...?

P(x)=?Skip Record If...? Q(x)=-1

由一阶线性方程的求解公式

?Skip Record If...?

=?Skip Record If...?

?Skip Record If...?

14 ?Skip Record If...?

两边同乘以?Skip Record If...??Skip Record If...?

令?Skip Record If...??Skip Record If...?

?Skip Record If...?这是n=2时的伯努利方程。

两边同除以?Skip Record If...??Skip Record If...?令?Skip Record If...?

?Skip Record If...??Skip Record If...?

P（x）=?Skip Record If...? Q(x)=?Skip Record If...?

由一阶线性方程的求解公式

?Skip Record If...?

=?Skip Record If...?

=?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

15?Skip Record If...??Skip Record If...?

?Skip Record If...??Skip Record If...?

?Skip Record If...?这是n=3时的伯努利方程。

?Skip Record If...?两边同除以?Skip Record If...??Skip Record If...?

令?Skip Record If...??Skip Record If...?

?Skip Record If...?=?Skip Record If...? P(y)=-2y Q(y)=?Skip Record If...?

由一阶线性方程的求解公式

?Skip Record If...?

=?Skip Record If...?

=?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

16 y=?Skip Record If...?+?Skip Record If...?

?Skip Record If...?

?Skip Record If...?

P(x)=1 Q(x)=?Skip Record If...?由一阶线性方程的求解公式

?Skip Record If...?

=?Skip Record If...?

=?Skip Record If...?

?Skip Record If...?

c=1

y=?Skip Record If...?

设函数?Skip Record If...?(t)于?Skip Record If...?∞

试求此函数。

令t=s=0 得?Skip Record If...?(0+0)=?Skip Record If...?(0)?Skip Record If...?(0) 即?Skip Record If...?(0)=?Skip Record If...?故?Skip Record If...?或?Skip Record If...?

（1）当?Skip Record If...?时?Skip Record If...?即?Skip Record If...?

?Skip Record If...?∞,?Skip Record If...?∞)

(2) 当?Skip Record If...?时?Skip Record If...?=?Skip Record If...?

=?Skip Record If...?=?Skip Record If...?

=?Skip Record If...?

于是?Skip Record If...?变量分离得?Skip Record If...?积分?Skip Record If...?

由于?Skip Record If...?，即t=0时?Skip Record If...? 1=?Skip Record If...??Skip Record If...?c=1

故?Skip Record If...?

20.试证：

（1）一阶非齐线性方程（2 .28）的任两解之差必为相应的齐线性方程（2.3）之解；

（2）若?Skip Record If...?是（2.3）的非零解，而?Skip Record If...?是（2.28）的解，则方程（2.28）的通解可表为?Skip Record If...?，其中?Skip Record If...?为任意常数.

（3）方程（2.3）任一解的常数倍或任两解之和（或差）仍是方程（2.3）的解.

证明：?Skip Record If...?（2.28）

?Skip Record If...??Skip Record If...?（2.3）

设?Skip Record If...?，?Skip Record If...?是（2.28）的任意两个解

则?Skip Record If...??Skip Record If...?（1）

?Skip Record If...?（2）

（1）-（2）得

?Skip Record If...?

即?Skip Record If...?是满足方程（2.3）

所以，命题成立。

由题意得：

?Skip Record If...?（3）

?Skip Record If...?（4）

1）先证?Skip Record If...?是（2.28）的一个解。

于是?Skip Record If...?得

?Skip Record If...?

?Skip Record If...?

故?Skip Record If...?是（2.28）的一个解。

2）现证方程（4）的任一解都可写成?Skip Record If...?的形式