Question

y'- $frac{2x-5}{ x^{2} }$*y=5 y(2)=4
найти решение задачи Коши

Answer 1

$y'- frac{2x-5}{ x^{2} } y=5\y=uv ;y'=u'v+v'u\u'v+v'u-uv*frac{2x-5}{ x^{2} }=5\u'v+u(vfrac{5-2x}{ x^{2} }+v')=5\begin{cases}vfrac{5-2x}{ x^{2} }+v'=0\u`v=5end{cases}\vfrac{5-2x}{ x^{2} }+frac{dv}{dx}=0|*frac{dx}{v}\frac{5x-2}{ x^{2} }dx=-frac{dv}{v}\intfrac{5-2x}{ x^{2} }dx=-intfrac{dv}{v}\5intfrac{dx}{x^2}-2intfrac{dx}{x}=-intfrac{dv}{v}\-frac{5}{x}-2ln|x|=-ln|v|\v=e^{frac{5}{x}+2ln|x|}=x^2e^{frac{5}{x}}\u'x^2e^{frac{5}{x}}=5\u`=5x^{-2}e^{-frac{5}{x}}$
$u=intfrac{5}{x^{2}}e^{-frac{5}{x}}dx=int e^{-frac{5}{x}}d(-frac{5}{x})=e^{-frac{5}{x}}+C\y=(e^{-frac{5}{x}}+C)*x^2e^{frac{5}{x}}=x^2(1+e^{frac{5}{x}}C)\y(2)=4\4=4(1+e^frac{5}{2}C)\1=1+e^{frac{5}{2}}\e^{frac{5}{2}}C=0\C=0\y=x^2$
Ответ:y=x²