Question

16. Решите уравнение: х" +x' =1, x(0) = 0, x'(0) =1​

Answer 1

Ответ:

$x''+x'=1\ \ ,\ \ \ x(0)=0\ ,\ x'(0)=1\\\\x=x(t)\ \ ,\ \ x'=p(t)\ \ ,\ \ x''=p'\ \ ,\\\\p'+p=1\ \ ,\ \ \ p'=1-p\ \ ,\ \ \ \dfrac{dp}{dt}=1-p\ \ ,\\\\\int \dfrac{dp}{1-p}=\int dt\ \ ,\ \ \ -ln|1-p|=t+C\ \ ,\ \ \dfrac{1}{1-p}=t+C_1\ \ ,\\\\\\1-p=\dfrac{1}{t+C_1}\ \ ,\ \ \ p=x'=1-\dfrac{1}{t+C_1}\ \ ,\ \ \dfrac{dx}{dt}=1-\dfrac{1}{t+C_1}\ \ ,\\\\\\\int dx=\int \Big(1-\dfrac{1}{t+C_1}\Big)\, dt\ \ ,\\\\\\\underline{\ x_{obshee}=t-ln|t+C_1|+C_2\ }$

$x(0)=0:\ \ \ 0=0-lnC_1+C_2\ \ ,\ \ \ C_2-lnC_1=0\ \ ,\ \ C_2=lnC_1\ ,\\\\x'(0)=1:\ \ x'=1-\dfrac{1}{t+C_1}\ \ ,\ \ 1=1-\dfrac{1}{C_1}\ \ ,\ \ \dfrac{1}{C_1}=0\ \ neverno\ !\ \ \Rightarrow$

Начальные условия заданы некорректно. Можно задать  $x'(0)=0$  .

$0=1-\dfrac{1}{C_1}\ \ ,\ \ \dfrac{1}{C_1}=1\ \ ,\ \ C_1=1\\\\\\C_2=lnC_1=ln1=0\\\\\\\underline{\ x_{chastnoe}=t-ln|\, t+1|\ }$