Question

Дифференциальное уравнение первого порядка ​

Answer 1

Ответ:

$(1+e^{x})\cdot y\cdot y'=e^{y}\ \ ,\ \ \ y(0)=0\\\\(1+e^{x})\cdot y\cdot \dfrac{dy}{dx}=e^{y}\\\\\int \dfrac{y\cdot dy}{e^{y}}=\int \dfrac{dx}{1+e^{x}}\\\\\\\int y\cdot e^{-y}\, dy=\Big[\ u=y\ ,\ dv=e^{-y}\, dy\ ,\ du=dy\ ,\ v=-e^{-y}\ \Big]=\\\\=uv-\int v\, du=-y\cdot e^{-y}+\int e^{-y}\, dy=-y\cdot e^{-y}-e^{-y}+C_1=-e^{-y}\cdot (y+1)+C_1\ ;\\\\\\\int \dfrac{dx}{1+e^{x}}=\Big[\ t=1+e^{x}\ ,\ e^{x}=t-1\ ,\ x=ln(t-1)\ ,\ dx=\dfrac{dt}{t-1}\ \Big]=$

$=\int \dfrac{dt}{(t-1)\cdot t}=\int\dfrac{dt}{t-1}-\int \dfrac{dt}{t}=ln|\, t-1|-ln|\, t|+C_2=\\\\\\=ln|\, e^{x}\, |-ln|\, 1+e^{x}\ |+C_2=ln\Big|\, \dfrac{e^{x}}{1+e^{x}}\, \Big|+C_2\ ;\\\\\\-e^{-y}\cdot (y+1)=ln\Big|\, \dfrac{e^{x}}{1+e^{x}}\, \Big|+C\\\\y(0)=0:\ \ \ -1=ln\dfrac{1}{2}+C\ \ ,\ \ -1=-ln2+C\ \ ,\ \ \ C=ln2-1\\\\\\-e^{-y}\cdot (y+1)=ln\Big|\, \dfrac{e^{x}}{1+e^{x}}\, \Big|+ln2-1\\\\\\\boxed{\ y=-e^{y}\cdot \Big(ln\Big|\, \dfrac{e^{x}}{1+e^{x}}\, \Big|+ln2-1\Big)-1\ }$