常微分方程(第三版)答案常微分方程习题答案2.11.«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 RecordIf...»«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 RecordIf...»=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。