Question

Дифференцированные уравнения

Answer 1

$1)\ \ \ y'+2xy=2x\, e^{-x^2}\ \ ,\ \ y(0)=5\\\\y=uv\ \ ,\ \ u'v+uv'+2xuv=2xe^{-x^2}\\\\u'v+u\, (v'+2xv)=2x\, e^{-x^2}\\\\a)\ \ \dfrac{dv}{dx}=-2xv\ \ ,\ \ \int \dfrac{dv}{v}=-2\int x\, dx\ \ ,\ \ ln|v|=-x^2\ \ ,\ \ v=e^{-x^2}\\\\b)\ \ \dfrac{du}{dx}\cdot e^{-x^2}=2x\, e^{-x^2}\ \ \int du=\int 2x\, dx\ \ ,\ \ u=x^2+C\\\\c)\ \ y=e^{-x^2}\, (x^2+C)\\\\d)\ \ y(0)=5:\ \ y(0)=e^0\cdot (0+C)=5\ \ ,\ \ C=5\\\\\underline {\ y=e^{-x^2}\, (x^2+5)\ }$

$2)\ \ y'-\dfrac{3y}{x}=e^{x}\cdot x^3\ \ ,\ \ y(1)=e\\\\y=uv\ ,\ \ u'v+uv'-\dfrac{3uv}{x}=e^{x}\cdot x^3\ \ ,\ \ u'v+u\, (v'-\dfrac{3v}{x})=e^{x}\cdot x^3\ \ ,\\\\a)\ \ \dfrac{dv}{dx}=\dfrac{3v}{x}\ \ ,\ \ \int \dfrac{dv}{v}=3\int \dfrac{dx}{x} \ \ ,\ \ ln|v|=3ln|x|\ \ ,\ \ v=x^3\\\\b)\ \ \dfrac{du}{dx}\cdot x^3=e^{x}\cdot x^3\ \ ,\ \ \int du=\int e^{x}\, dx\ \ ,\ \ u=e^{x}+C\\\\c)\ \ y=x^3(e^{x}+C)\\\\d)\ \ y(1)=e:\ \ y(1)=e+C=e\ \ ,\ \ C=0\\\\\underline {\ y=x^3\, e^{x}\ }$